Questions tagged [muirhead-inequality]
Inequality proof by using the Muirhead inequality.
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Question on the Proof of Muirhead's Inquality (cited from AoPS)
In AoPS' (Art of Problem Solving) proof of Muirhead's inequality, how does the below equality work out?
The below equation appears to show two expressions (1 and 2), each under the symmetric sum ...
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4answers
84 views
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+...
2
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1answer
71 views
For any real positive numbers $a, b, c$, prove that $3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq abc(a+b+c)^3$ [duplicate]
My progress is that I applied Hölder’s for this,
$3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} \geq (a+b+c)^3$
whereas $3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} = (1+1+1)(a^2b+b^2c+c^2a)(\...
1
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3answers
182 views
Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\...
2
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6answers
125 views
How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$?
Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
...
3
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3answers
116 views
Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$
I was trying this question with factorization and other similar methods,
Let $a, b, c \geq 0$. Prove that
$$\begin{array}{c}
\sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt]
\...
3
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2answers
162 views
AM/GM inequalities
I need some help to prove this inequality... I guess one can use Jensen's then AM/GM inequalities.
Let $x_1, x_2, x_3, x_4$ be non- negative real numbers such that
$x_1 x_2 x_3 x_4 =1$.
We want to ...
2
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1answer
76 views
Proof of inequality by Muirhead
We have to prove:
$$\frac{\sqrt{pq}}{p+q+2r}+\frac{\sqrt{pr}}{p+r+2q}+\frac{\sqrt{pr}}{p+r+2q}\leq\frac{3}{4}$$
By multiplying it all out we get the following equivalent:
\begin{align*}
4\sum_{cyc}{...
1
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3answers
86 views
Inequality 6 deg
For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+...
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0answers
127 views
Prove that if $abc \geq 1$ and $a,b,c > 0$ then $\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$ [duplicate]
Prove that if $abc \geq 1$ and $a,b,c > 0$ then,
$$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$$
Can anyone help me with this problem?
Tried AM-GM and Cauchy-Schwarz ...
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2answers
55 views
Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$
For $a,b,c \in (0,1)$ such that $ab+bc+ca=1$ Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$
I tried repleace $1$ by $ab+...
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2answers
71 views
Inequality involving AM-GM but its wierd [duplicate]
Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{...
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3answers
108 views
Inequality question . [duplicate]
Let $a,b,c>0$ with$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that
$(a + 1)(b + 1)(c + 1) \geq 64$
Ohk so we are given that $abc=a+b+c$ with that now the inequality becomes $2abc+(a+b+c)+1 \...
3
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2answers
77 views
Inequality question.
Let $a,b,c>0$ with $a+b+c=1$. Show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{\left(a^3 + b^3 + c^3\right)}{abc}$$
Ohhhkk. So first off,
\begin{align} a^3 + b^3+ c^3 &...
2
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1answer
76 views
Schur inequality
Show that for all positive real numbers $a$, $b$ and $c$ such that $abc=1$, the inequality $a+b+c+2a^4+2b^4+2c^4\ge \dfrac{3}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\...
5
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3answers
81 views
Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I ...
1
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1answer
62 views
Stuck when transforming and solving this
Given abc=1 ( all positive real numbers). Prove that:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$
My attempt:
$$\frac ab + \frac ...
1
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1answer
129 views
Proving $\sum\limits_{\rm cyc} 1/(a^2 -b+4) \geq3/4$
Suppose $a,b,c\in\mathbb R^+$ with $a+b+c=3.$ Prove that $$\frac{1}{a^2- b+4}+\frac{1}{b^2-c+4}+\frac{1}{c^2- a+4}\geqslant\frac{3}{4}.$$
I tried various approaches, but nothing seems to work. ...
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2answers
66 views
Prove $13\sum(a+b)^5 \geq 16 \sum ab(a+b)(4a^2+4b^2+4ab+c^2)$
Given $a,b,c>0$, prove that $$13[(a+b)^5+(b+c)^5+(c+a)^5] \geq 16[ab(a+b)(4a^2+4b^2+4ab+c^2)+bc(b+c)(4b^2+4c^2+4bc+a^2)+ca(c+a)(4c^2+4a^2+4ca+b^2)]$$
I tried subtracting the RHS from the LFS but ...
3
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2answers
415 views
Barnard and Child inequality exercise
Prove that,
$$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)≥abc(a+b+c)^3$$
For positive $a,b,c$
The exercises in this book are making me crazy.
Any help would be appreciated.
My attempts:
I opened the LHS ...
0
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1answer
129 views
Another asymmetric inequality $5+\frac{3(a^2+2b^2+c^2)}{(a+b)(b+c)} \geq \frac{9(a+b)(b+c)(c+a)}{(a+b+c)(ab+bc+ca)}$
A while ago I conjectured this inequality and its (little) sister on AOPS. Here is another related inequality in the opposite direction which I strongly suspect is true, although I don't have a proof:
...
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2answers
71 views
Prove that for all positive real numbers $a,b,$ and $c$ we have $a^n+b^n+c^n \geq a^{n-2}bc+ab^{n-2}c+abc^{n-2}$.
Let $a,b,$ and $c$ be positive real numbers.Prove:$$a^n+b^n+c^n \geq a^{n-2}bc+ab^{n-2}c+abc^{n-2}$$
n is a positive integer greater than 2.
Icould solve this using Jensen's inequality. I am ...
2
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1answer
80 views
How to prove $a^3+b^3+c^3+d^3+3\left(a+b+c+d\right) \geq 14+2abcd$ when $a^2+b^2+c^2+d^2=4$
This is a problem from AoPS I can't solve:
Let $a,b,c,d\geq0$ with $a^2+b^2+c^2+d^2=4$. How can I prove: $$a^3+b^3+c^3+d^3+3\left(a+b+c+d\right) \geq 14+2abcd$$
My attempt:
I try setting $a=2\cos(x)...
1
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1answer
55 views
For real numbers $x>0, y>0, z>0$ and $x y z=1 .$ prove that $ x^{6}+y^{6}+z^{6} \geq x^{5}+y^{5}+z^{5} $
For real numbers $x>0, y>0, z>0$ and $x y z=1 .$ Prove that $$
x^{6}+y^{6}+z^{6} \geq x^{5}+y^{5}+z^{5} $$
How to show this?
I tried using https://en.wikipedia.org/wiki/Muirhead%...
3
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2answers
86 views
Proof verification for $x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9$ (where $xyz=1$ and $x,y,z\in \mathbb{R}^+$)
My teacher has shown me the following problem:
Problem. Let $x,y,z\in \mathbb{R}_+$ with $xyz=1$. Show that:$$x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9.$$
I think I solved the problem using Muirhead's ...
2
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2answers
268 views
Proving $\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{c}{c^2+2a^2}\geq 1$ when $a^2+b^2+c^2=a^3+b^3+c^3$
In this answer, @MichaelRozenberg stated the following inequality:
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=a^2+b^2+c^2.$ Then
$$\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{...
2
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4answers
170 views
If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$
The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by ...
1
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2answers
44 views
Prove an inequality : $\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$ without Jensen's inequality
I'm interested in the following problem :
Let $a,b,c>0$ be the variables and $u,v>0$ be constant then we have :
$$\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$$
Rewrrting the ...
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3answers
94 views
Prove that $3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$.
Let $a,b\geq 0$. How can I prove $$3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$$
?
I try using Schur and Muirhead but they didn't work here because Schur is for three variables and Muirhead ...
3
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3answers
60 views
Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$
Hello ladies and gentlemen, here I have another inequality that I am struggling with:
Let $a,b,c>0$ Then
$$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$
I try to show $$\frac{ab}{ab+b^2+ca}\le\...
4
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1answer
157 views
Inequality in 4 variables Vasc's EV
Let $a,b,c,d\geq0$ satisfying $a+b+c+d=4$ . Prove
$$\sqrt{a^3+b^3+c^3+d^3}+2(\sqrt3
-1)abcd\geq\sqrt{3(abc+abd+acd+bcd)}$$
Attempt: $a^3+b^3+c^3+d^3=(a+b+c+d)(a^2+b^2+c^2+d^2-ab-bc-cd-da-ac-bd)+3(...
2
votes
4answers
123 views
How do we prove this inequality?
Suppose $a,b,c > 0$. Prove that
$$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$
I've tried multiplying everything by the denominator and then I tried to ...
2
votes
3answers
203 views
Problematic inequality & hint
I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried ...
0
votes
3answers
123 views
Generalization of the AM-GM inequality for three variables
Theorem. Let $a,b,c$ be three non-negative real numbers. Then $$a^6+b^6+c^6\geq 3a^2b^2c^2+\frac12 (a-b)^2 (b-c)^2 (c-a)^2.$$
Remark. This Theorem is a generalization of the AM-GM inequality for ...
4
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3answers
195 views
A hard inequality indian olympiad problem [duplicate]
If $x,y,z$ are positive real numbers, prove that:
$\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$.
I have been ...
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1answer
49 views
An inequality with four variables
Prove that
$$4[(4;0;0;0)]+12[(2;1;1;0)]\ge 12[(3;1;0;0)]+3[(1;1;1;1)]$$
with $$[(a;b;c;d)]=\frac{1}{4!}\sum_{\sigma\in Sym(4)}x_{\sigma(1)}^{a}\cdot x_{\sigma(2)}^b\cdot x_{\sigma(3)}^c\cdot x_{\...
1
vote
4answers
150 views
How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? [duplicate]
There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:
If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$
...
5
votes
4answers
172 views
How can we not use Muirhead's Inequality for proving the following inequality?
There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to ...
1
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1answer
49 views
an interesting inequality with Muirhead
If $x,y,z>0$ I have to prove that
$\sum\limits_{cyc}^{} \frac { x(x^3 yz+x^2-x y^3 z-yz) }{(1+x y^2)(1+xyz)} \ge 0$ holds. My approach is that from Muirhead's inequality the inequality is true ...
5
votes
1answer
146 views
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .
Ji Chen. Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{cyc}\sqrt{\frac{a+ b}{b+ 1}}\geqq 3\sqrt[3]{\frac{4\,abc}{3\,abc+ 1}}$$
Of course, we've to solve it by $uvw$, before that,...
5
votes
3answers
126 views
Prove $x+y+z \ge xy+yz+zx$
Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that
$$x+y+z \ge xy+yz+zx$$
My try:
Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get
$$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$
$$-(a+b+c)-(a+b+c)+ab+bc+...
2
votes
2answers
97 views
Calculate the maximum value of $\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$ where $x, y, z > 0$ and $x^2 + y^2 + z^2 = 3xyz$.
$x$, $y$ and $z$ are positives such that $x^2 + y^2 + z^2 = 3xyz$. Calculate the maximum value of $$\large \frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$$
This is (obviously) ...
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1answer
95 views
Nice refinement of an inequality by Michael Rozenberg
It's related to this If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$:
In fact we have this refinement (wich I think much easier) :
Let $a,b,c>0$ then we have :
$$\...
2
votes
1answer
149 views
show this inequality with $xy+yz+zx=3$
let $x,y,z>0$ and such $xy+yz+zx=3$,show that
$$\dfrac{x}{x^3+y^2+1}+\dfrac{y}{y^3+z^2+1}+\dfrac{z}{z^3+x^2+1}\le 1$$
To prove this inequality,I want use following Cauchy-Schwarz inequality
$$(x^3+...
0
votes
3answers
151 views
Find $f(x)$ so that $\sum\limits_{cyc}a^{2}-f(x)\left(\prod_{sym}a-\prod_{sym}(1-a)\right)\geqq3(\frac{x}{2})^{2}$ .
Find $f(x)$ so that
$$\sum\limits_{cyc}a^{2}- f(x)\left (\prod\limits_{sym}a- \prod\limits_{sym}(1- a) \right )\geqq 3\left ( \frac{x}{2} \right )^{2}$$
OP. Given three numbers $a, b, c$ so that $\{...
0
votes
2answers
61 views
A three variable inequality doubt , can I consider the three variables into just one variable , and show the inequality.
I was trying to prove the inequality : for a,b,c positive real numbers where $abc=1$ prove $$\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}}\leq 1 . $$
It is easy ...
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1answer
115 views
Inequality with $a+b+c=1$, a,b,c positive numbers
If a,b,c are real positive numbers, such as $ a+b+c=1$ prove that $ \left( ab+bc+ca\right) \left( 1+3abc\right) \geq 10abc $.
Thank you!
First try:$$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(...
0
votes
2answers
102 views
Show that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\leq 1$.
Let $a, b, c>0$ s.t. $abc (a+b+c)=3$.
Show that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+ \frac{1}{c^2+a^2+1}\leq 1$.
I have no idea how to start.
-1
votes
1answer
117 views
Show that $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq 256$. [duplicate]
Let $a, b, c, d\geq 0$ s.t. $a+b+c+d=4$.
Show that $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq 256$.
I don't know how can I deconditioned the inequality.
0
votes
0answers
55 views
Inequality using lengths of the edges of a triangle
If $a,b,c $ are the lengths of the edges of a triangle, show that:
$$\frac {6 (a^2+b^2+c^2)}{a+b+c}\geq \frac {(a+b)^2}{b+c}+\frac {(b+c)^2}{a+c}+\frac {(c+a)^2}{a+b} $$
I have no idea how to start.
