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Questions tagged [uvw]

This is a very useful method for the proof of polynomial inequalities with three variables. Sometimes it works for more variables.

-3
votes
1answer
51 views

Prove $\sum\limits_{cyc}\frac{ab}{b^{\,2}+ c^{\,2}}\geqq \frac{3}{2}$

For $a\geqq b\geqq c> 0$. Prove $$\frac{ab}{b^{\,2}+ c^{\,2}}+ \frac{bc}{c^{\,2}+ a^{\,2}}+ \frac{ca}{a^{\,2}+ b^{\,2}}\geqq \frac{3}{2}$$ I used discriminant to find & want to see a solution ...
1
vote
0answers
179 views

Prove $\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$ for $x,\,y,\,z\in [\,j,\,k\,j\,]$

For $x,\,y,\,z\in [\,j,\,k\,j\,]$ $$\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$$ is true with $j= constant,\,k= constant> 8$ and $k_{\,\max}$...
0
votes
3answers
125 views

Prove $1+ a^{\,2}+ b^{\,2}+ c^{\,2}+ 4\,abc\geqq a+ b+ c+ ab+ bc+ ca$ for $a,\,b,\,c\geqq 0$

Given that $a,\,b,\,c$ are $3$ non-negatve numbers, prove $$1+ a^{\,2}+ b^{\,2}+ c^{\,2}+ 4\,abc\geqq a+ b+ c+ ab+ bc+ ca$$ Let $X= a+ b+ c$$,$ we have to prove $$\left ( \frac{1}{X^{\,3}}- \frac{1}{X^...
1
vote
2answers
50 views

Prove $2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$ [on hold]

For $a,\,b,\,c\geqq 0$ and $b\equiv {\rm mid}\,\{\,a,\,b,\,c\,\}$. Prove $$2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$$ Inspried from $\lceil$ Prove $k=0$ is the only non-...
1
vote
1answer
76 views

Prove $F(\,k\,)=(\,16\,X^{\,}- 24\,X+ 18\,)k^{\,2}- 11\,X+ 1\geqq 0$

Given that $1\leqq X\leqq k,$ prove that $$F(\,k\,)=(\,16\,X^{\,}- 24\,X+ 18\,)k^{\,2}- 11\,X+ 1\geqq 0 \tag{29}$$ Origin For $a,\,b,\,c\geqq 0$ and $a+ b+ c= 3$, prove that $(2+a^2)(2+b^2)(2+c^2)+...
0
votes
3answers
93 views

How to find $f(X)$ such that $\sum\limits_{cyc}a^2-f(X)[abc-(1-a)(1-b)(1-c)]\geqq\frac{3}{4}X^2$ for $abc=(X- a)(X- b)(X- c),0\leqq a,\,b,\,c\leqq X$?

We have $\sum\limits_{cyc}\,a^{\,2}\geqq \frac{3}{4}\,X^{\,2}\tag{HaiDangel29}$ with $abc= (\,X- a\,)(\,X- b\,)(\,X- c\,),\,0\leqq a,\,b,\,c\leqq X$. Here is a hint to get you started from above. For $...
0
votes
2answers
94 views

Prove/disprove $\sum_{cyc}a\sqrt{\frac{(ca + 1)(ab + 1)}{bc + 1}} \ge 2$ where $a$, $b$, $c > 0$ and $a^2 + b^2 + c^2 = 1$

$a$, $b$ and $c$ are positives such that $a^2 + b^2 + c^2 = 1$. Prove/disprove that $$a\sqrt{\frac{(ca + 1)(ab + 1)}{bc + 1}} + b\sqrt{\frac{(ab + 1)(bc + 1)}{ca + 1}} + c\sqrt{\frac{(bc + 1)(ca + 1)}{...
0
votes
1answer
85 views

Prove $\frac{1}{a+ 2\,b}+ \frac{1}{b+ 2\,c}+ \frac{1}{c+ 2\,a}\leqq 1$ with $8\,abc\geqq a+ b+ c+ 5$ and $a,\,b,\,c> 0$

Prove $$\frac{1}{a+ 2\,b}+ \frac{1}{b+ 2\,c}+ \frac{1}{c+ 2\,a}\leqq 1$$ with $8\,abc\geqq a+ b+ c+ 5$ and $a,\,b,\,c> 0$ $$constant= 8$$ is the best $constant$, which was found by me (using ...
0
votes
1answer
110 views

For $a,b,c>0,\,a+b+c=3,$ prove $\sum\limits_{cyc}\,\frac{1}{a}\geqq\left(\frac{1}{2}+\frac{5}{18}\,\sqrt{3}\right)(a^2+b^2+c^2)$

Given $a,\,b,\,c> 0$ such that$:$ $a+ b+ c= 3$$.$ Prove$:$ $$\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}\geqq \left ( \frac{1}{2}+ \frac{5}{18}\,\sqrt{3} \right )(\,a^{\,2}+ b^{\,2}+ c^{\,2}\,)$$ I find $...
0
votes
0answers
60 views

Prove $\sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,a+ b\,)}}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,c+ a\,)}}$ with $a,\,b,\,c> 0$

Let $a,\,b,\,c$ be positive numbers. Prove that $$\sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,a+ b\,)}}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,c+ a\,)}}$$ I tried Holder and $\lceil$ https://...
0
votes
0answers
65 views

Prove that $\sum\limits_{cyc}\,\frac{a^{\,2}}{bc+ a}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{2\,bc+ 2}}$ [closed]

Let $a,\,b,\,c$ be positive numbers. Prove that $$\sum\limits_{cyc}\,\frac{a^{\,2}}{bc+ a}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{2\,bc+ 2}}$$ I tried Holder Inequality (it's only the hint to get you ...
8
votes
3answers
668 views

Hard inequality $ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $

I need to prove or disprove the following inequality: $$ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $$ For $x,y,z \in \mathbb R^+$. I found no counter ...
0
votes
1answer
99 views

Prove $\prod\,\left ( a+ \frac{1}{a} \right )- \frac{4}{3}\sum\,\frac{b+ c}{a}\geqq 0 $

Prove $$\begin{equation}\begin{split} \prod\,\left ( a+ \frac{1}{a} \right )- \frac{4}{3}\sum\,\frac{b+ c}{a}\geqq 0 \end{split}\end{equation}$$ with $a,\,b,\,c> 0$. $$\begin{equation}\begin{...
2
votes
1answer
89 views

show this $\sum_{cyc}\frac{x}{x^2-x+1}\le\frac{8}{3}$ [duplicate]

let $x,y,z,w\in R$,and such $x+y+z+w=2$.show that $$\sum_{cyc}\dfrac{x}{x^2-x+1}\le\dfrac{8}{3}$$ I have only solve when $x,y,z,w>0$, because $$\dfrac{x}{x^2-x+1}\le\dfrac{4}{3}x$$ so $$\sum_{...
1
vote
2answers
95 views

Find maximum of function $A=\sum _{cyc}\frac{1}{a^2+2}$

Let $a,b,c\in R^+$ such that $ab+bc+ca=1$. Find the maximum value of $$A=\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}$$ I will prove $A\le \dfrac{9}{7}$ and the equality occurs when $a=b=c=\dfrac{...
12
votes
3answers
293 views

Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$ If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ ...
20
votes
4answers
902 views

How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$

let $a,b,c>0$,and such $a+b+c=3$, show that $$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$ I think this inequality use this $$ab\le\dfrac{(a+b)^2}{4}$$
6
votes
4answers
329 views

Prove this trigonometric inequality about the angles of $\triangle ABC$

In $\Delta ABC$ show that $$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$ since $$\frac{\sqrt{3}}{2}\...
1
vote
2answers
74 views

Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$

Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of $$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$ To find the minimum of $P$ ...
4
votes
1answer
254 views

Prove that $\sum_{cyc}\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}\ge \frac{\sqrt{181}}{5}$

For $a,b,c>0$ satisfy $ab+bc+ca\ge \frac{4}{3}$. Prove that $$\sqrt{a^2+\frac{1}{\left(b+1\right)^2}}+\sqrt{b^2+\frac{1}{\left(c+1\right)^2}}+\sqrt{c^2+\frac{1}{\left(a+1\right)^2}}\ge \frac{\sqrt{...
2
votes
3answers
62 views

Inequalities, but working around an absolute value

So what I want to prove is $$ |xy+xz+yz- 2(x+y+z) + 3| \leq |x^2+y^2+z^2-2(x+y+z)+3| $$ for $x,y,z\in \mathbb{R}$, and I'm aware that the RHS is just $|(x-1)^2+(y-1)^2+(z-1)^2|$. Now I'm able to ...
2
votes
2answers
76 views

Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$

Let $a,b,c$ be reals with $a+b+c=1$. Show that : $$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$ I have tried to something like this: $$18a^4-24a^3+6a^2-12a+12\geq 0$$ $$18b^4-24b^3+6b^2-12b+...
3
votes
3answers
133 views

Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$

Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that: $$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$ My try We have: $$\left ( a+ b \...
6
votes
3answers
322 views

How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$

let $a,b,c\ge 0$, such that $a+b+c=1$, prove that $$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$ This problem is simple as 2005, china west competition problem $$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge ...
4
votes
3answers
300 views

$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$?

For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ I checked in very many cases. Example :$c=1, a=2,b=\frac{1}{2}...$ then it’s ...
5
votes
2answers
257 views

Prove $\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$

Let $x,y,z>0$. Prove that $$\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$$ On the left-hand side we have arithmetic and harmonic means, while on the right ...
-1
votes
2answers
65 views

Find min of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+4\sqrt{2}\sqrt{\frac{ab+bc+ac}{a^2+b^2+c^2}}$ [closed]

Given the three real numbers a, b, c are not negative, in which at most some are equal to zero. Find min of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+4\sqrt{2}\sqrt{\frac{ab+bc+ac}{a^2+b^2+c^2}}$ ...
1
vote
1answer
117 views

Square root inequality cyclic

Is it true that if $a+b+c=1$, where $a$, $b$ and $c$ are nonnegative real numbers, then $$(3-2\sqrt{2})\sum\limits_{cyc}\sqrt{ab}+2\sqrt{2}-1\geq\sum\limits_{cyc}\sqrt{(1-a)(1-b)}?$$ Edit: I was told ...
3
votes
1answer
83 views

Prove $\Sigma_{cyc}(\frac{a}{b-c}-3)^4\ge193$

The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following ...
2
votes
1answer
206 views

Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $

Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds: $$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$ What I ...
2
votes
2answers
499 views

How to prove $\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$

Question: let $a,b,c>0$,show that: $$\sqrt{\frac{ab}{2a^2+bc+ca}}+\sqrt{\frac{bc}{2b^2+ca+ab}}+\sqrt{\frac{ca}{2c^2+ab+bc}}\ge\frac{81}{2}\cdot\frac{abc}{(a+b+c)^3}$$ maybe this inequality ...
5
votes
4answers
199 views

Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$. [duplicate]

Find the maximum value of the expression : $${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$ where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$. Taking $x=y=z=\...
0
votes
0answers
224 views

Prove that $\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{1}{ab+bc+ca}$ [duplicate]

For $a,b,c>0$. Prove that $$\frac{1}{\left(2a+b\right)^2}+\frac{1}{\left(2b+c\right)^2}+\frac{1}{\left(2c+a\right)^2}\ge\frac{1}{ab+bc+ca}$$ Outside $a=b=c$ I can't exploit what from it
7
votes
4answers
381 views

Prove $\sum{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$

Let $x,y,z>0$: Prove that: $\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}$ I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
0
votes
1answer
129 views

Prove $(x+y+z)^8\ge 3^7(x^2+y^2+z^2)x^2y^2z^2$

Let $x,y,z>0$ show that $$(x+y+z)^8\ge 3^7(x^2+y^2+z^2)x^2y^2z^2$$ Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them. Things ...
5
votes
2answers
235 views

Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$

The following inequality is exercise 1.8 from this book. For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$ I've managed to prove this via brut-...
2
votes
6answers
97 views

Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$

Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$. I expanded the brackets and applied AM-GM on all of the eight terms to get : $$\big(3+2a^2\big)\big(3+...
2
votes
2answers
147 views

Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$

How one can prove the following. Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds: $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$ WLOG one can assume that $0\le a\le b\le c$. ...
2
votes
4answers
1k views

Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $

Let $x,y,z > 0$ and $xyz=8.$ Prove that $$ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $$ I have tried with AM–GM inequality but no result.
0
votes
2answers
76 views

Non - symetric inequality in 3 variables

How can we prove the following inequality for $a,b,c>0$: $$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}\leq \dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{c^2+a^2} \ ?$$
9
votes
2answers
181 views

I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$

Let $x,y,z>0$,prove or disprove $$\sqrt[4]{\dfrac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\dfrac{(x+y)(y+z)(z+x)}{8}}$$ I tried many times ,use $$x^2+y^2+z^2\ge\dfrac{1}{3}(x+y+z)^2$$ and kown $$9(...
1
vote
3answers
113 views

Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$

Let $a,b,c>0; a+b+c=1$. Prove the inequality $$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$ My work so far: I tried AM-GM and used fact $a+b+c=1$.
0
votes
3answers
139 views

How to prove $|a^{2} + b^{2} + c^{2} - 2\left( ab+bc+ac\right)| \le \frac{1}{27}$

Let $a,b,c$ are $3$ edge of a triangle and $\sqrt{a} + \sqrt{b} + \sqrt{c} = 1.$ How to prove $|a^{2} + b^{2} + c^{2} - 2\left( ab+bc+ac\right)| \le \frac{1}{27}$? Can this be proved with simple ...
2
votes
2answers
129 views

Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$

Let $a,b,c>0$ prove or disprove $$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$ since $$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+a^2c^2)\tag{1}$$ other $$(a+...
2
votes
2answers
111 views

Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$

I can't make progress with proving this inequality. I have tried opening the brackets and using $abc=1$ in order to obtain the following: $$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^...
4
votes
6answers
231 views

Inequality with $a+b+c=1$

Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$ I am trying to resolve this problem but actually i found some ...
1
vote
1answer
71 views

How prove this inequality with $\sum\sqrt{a+b}\sum\frac{1}{\sqrt{2a+b+c}}\le\frac{9}{\sqrt{2}}$?

I found the following inequality which I can't solve it. Let $a,b,c>0$,prove or disprove $$(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})\left(\dfrac{1}{\sqrt{2a+b+c}}+\dfrac{1}{\sqrt{2b+c+a}}+\dfrac{1}{\...
2
votes
1answer
164 views

Prove $(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$

Let $a,b,c,d>0$ and such $a+b+c+d=1$, show that $$(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$$ since $$a+b\ge 2\sqrt{ab}$$ I think this will not hold.because we have $256abcd\le 1$
8
votes
5answers
523 views

Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$

Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$ It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use ...
4
votes
2answers
211 views

Inequality exercise (olympiad)

For positive $a$, $b$, $c$ such that $abc=1$. Show that $$(ab+bc+ca)(a+b+c)+6\geq 5(a+b+c).$$ From the LHS, using AM-GM, we see that $(ab+bc+ca)(a+b+c)+6\geq 3(abc)^{2/3}3(abc)^{1/3}+6=15$. But ...