5
votes
Accepted
How to prove $\frac{1}{\sqrt{3a^2+13b}}+\frac{1}{\sqrt{3b^2+13c}}+\frac{1}{\sqrt{3c^2+13a}} \ge \frac{3}{4}$?
The inequality $\sum\limits_{cyc}\sqrt{3a^2+13b}\leq12$ is wrong.
Try $c=0$ and $a=b=\frac{3}{2}.$
Bacteria helps!
By Holder:
$$\left(\sum_{cyc}\frac{1}{\sqrt{3x^2+13y}}\right)^2\sum_{cyc}(3a^2+13b)(...
- 197k
5
votes
How to prove $\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$
Here is my solution.
Step 1
Substitute $x=\frac{b+c}{\sqrt{bc}}$, $x=\frac{c+a}{\sqrt{ca}}$, $z=\frac{a+b}{\sqrt{ab}}$
Then the problem changes as below.
$$ \sqrt{\sum_{cyc} x^2 - 3} \geq 4 - \frac{4}{...
- 95
4
votes
Prove $\frac{1}{\sqrt{2a+b+c}}+\frac{1}{\sqrt{2b+a+c}}+\frac{1}{\sqrt{2c+b+a}}\le \frac{9}{2}\cdot\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$
Another way(iura).
Let $f(x)=3\sqrt{x}+\frac{2}{\sqrt{1+x}},$ where $x\in(0,1).$
Thus, $f$ is a concave function and by Jensen we obtain:
$$3(\sqrt{a}+\sqrt{b}+\sqrt{c})+2\sum_{cyc}\frac{1}{\sqrt{1+a}}...
- 197k
4
votes
$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$
Sketch of a proof.
We use the so-called isolated fudging.
It suffices to prove that
$$\frac{2+\sqrt{ab}}{\sqrt{ab}+c} + \frac{a^2 + b^2}{16abc} \ge \frac{39}{8}\cdot \frac{6a + 6b + abc}{12(a + b + c) ...
- 36.4k
4
votes
Accepted
closed-form expression for solution of inequality $ \left\lceil \frac{a}{x} \right\rceil > b$
$\lceil y\rceil >b$ is equivalent to $y>\lfloor b\rfloor,$ so your inequality is equivalent to $\frac ax>\lfloor b\rfloor.$ Since $a,b>0,$ this means $x>0,$ and thus the answer is $0<...
- 176k
4
votes
Accepted
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Some thoughts.
By Holder inequality, we have
\begin{align*}
&\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a^2 + 4bc}} \right)^2\cdot \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3 \\
\...
- 36.4k
4
votes
Prove $2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a)$ for $abc=1.$
Symmetrization helps.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u=xw$.
Thus,by AM-GM $x\geq1$ and we need to prove that
$$u\left(2w^3+2\sum_{cyc}a^2c\right)\geq(9uv^2-w^3)w$$ or
$$u\...
- 197k
4
votes
prove that $f’(x)e^{\lambda x}$ is increasing if and only if $f’(x)+\lambda f(x)$ is increasing. Where $f\in C^1(0,\infty)$.
I have decided to undelete this partial answer (it was a bit incorrect) but I hope the following observations contribute something. We have two statements:
i) $f'(x)e^{\lambda x}$ is increasing
ii) $f'...
- 496
4
votes
Show that $ I = \int_0^1 \frac{dx}{\sqrt{\sin(2x) + \sin(x) + 3}} < \frac{1}{2} $?
Using Taylor's Theorem we find that
$$\frac1{\sqrt{\sin(x)+\sin(2x)+3}}\le \frac1{\sqrt 3}-\frac{x}{2\sqrt 3}+\frac{\sqrt 3 x^2}{8}-\frac{x^3}{16\sqrt 3}$$
The proof of this bound is left to the ...
- 177k
4
votes
Accepted
How to prove $x^{x/(1+x)}/(1+x)\geq1/2$
By Bernoulli $$\frac{x^{\frac{x}{x+1}}}{1+x}=\frac{1}{(1+x)\left(1+\frac{1}{x}-1\right)^{\frac{x}{x+1}}}\geq\frac{1}{(1+x)\left(1+\left(\frac{1}{x}-1\right)\cdot\frac{x}{x+1}\right)}=\frac{1}{2}.$$
- 197k
3
votes
Accepted
Prove $\sqrt[3]{3a+bc}+\sqrt[3]{3b+ca}+\sqrt[3]{3c+ab}\le \frac{3}{2}\sqrt[3]{a^2+b^2+c^2+29}$ for $ab+bc+ca=3.$
By Holder
$$\left(\sum_{cyc}\sqrt[3]{3a+bc}\right)^3\leq\sum_{cyc}\frac{3a+bc}{(7a+5b+5c)^2}\left(\sum_{cyc}(7a+5b+5c)\right)^2$$ and it's enough to prove $$\sum_{cyc}\frac{3a+bc}{(7a+5b+5c)^2}\leq\...
- 197k
3
votes
Accepted
Finding $\small{\min\limits_{a+b+c=2}\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}.}$
Let $c=\min\{a,b,c\}$.
We'll prove that
$$\sqrt{a-2bc+3}+\sqrt{b-2ac+3}\geq\sqrt{2(a+b)-4c(a+b)+12-\frac{1}{2}(a-b)^2}.$$
Indeed, we need to prove that:
$$2\sqrt{(a-2bc+3)(a-2ac+3)}\geq a+b-2c(a+b)+6-...
- 197k
3
votes
$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$
Proof.
We can use the isolated fudging idea.
Indeed, we will prove $$\frac{2+\sqrt{bc}}{\sqrt{bc}+a}\ge 2-\frac{2a}{a+b+c+bc}. \tag{1}$$
Taking cylic sum on $(1)$ it remains to prove that$$\frac{a^2+b^...
- 1,874
3
votes
Accepted
Finding $\small{\min\limits_{a+b+c=1}P=\sqrt{2a^3+abc}+\sqrt{2b^3+abc}+\sqrt{2c^3+abc}. }$
By Holder
$$\left(\sum_{cyc}\sqrt{2a^3+abc}\right)^2\sum_{cyc}\frac{a^2}{2a^2+bc}\geq(a+b+c)^3$$ and it's enough to prove that:
$$\sum_{cyc}\frac{a^2}{2a^2+bc}\leq1,$$ which is true by C-S.
- 197k
3
votes
Accepted
Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}$ when $a+b+c=3.$
Let $a^2+b^2+c^2=x(ab+ac+bc).$
Thus, $x\geq1$ and by C-S and Muirhead we obtain:
$$\sum_{cyc}\sqrt{a+b}=\sqrt{2(a+b+c)+2\sqrt{(a+b)(a+c)}}=$$
$$=\sqrt{6+2\sqrt{\sum_{cyc}\left(a^2+3ab+2(a+b)\sqrt{(a+c)...
- 197k
3
votes
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Some thoughts.
By AM-GM, it suffices to prove that
$$\frac{2}{\frac{a^2 + 4bc}{2 + bc/2} + (2 + bc/2)} + \frac{2}{\frac{b^2 + 4ca}{2 + ca/2} + (2 + ca/2)} + \frac{2}{\frac{c^2 + 4ab}{2 + ab/2} + (2 + ...
- 36.4k
3
votes
Accepted
Prove $2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a)$ for $abc=1.$
Proof.
Firstly, rewrite the inequality as: $$2(a+b+c)(ab+bc+ca)(a^2c+c^2b+b^2a+1)\ge3(a+b)(b+c)(c+a)(ab+bc+ca)$$, $$\iff \frac{1+a^2c+c^2b+b^2a}{(a+b)(b+c)(c+a)}+1+a^2c+c^2b+b^2a\ge\frac{3(ab+bc+ca)}{...
- 1,874
3
votes
Accepted
Prove $\frac{1}{\sqrt{a+b+7c}}+\frac{1}{\sqrt{c+b+7a}}+\frac{1}{\sqrt{a+c+7b}}\ge 1.$
Proof.
By AM-GM, it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{2}{\frac{a + b + 7c}{2 + c} + (2 + c)} \ge 1. \tag{1}$$
We use the pqr method. Let $p = a + b + c, q = ab + bc + ca, r = abc$.
...
- 36.4k
3
votes
Prove $\frac{4}{(a+1)(b+1)(c+1)}+\frac{1}{4}\ge \frac{a}{(a+1)^2}+\frac{b}{(b+1)^2}+\frac{c}{(c+1)^2}.$
Here is what I thought. First,
$$(a-1)(b-1)(c+1) = abc + ab - ac - bc + c - a - b + 1$$
then
$$\sum_{cyc} (a-1)(b-1)(c+1) = 3abc + 3 - \sum_{cyc} ab - \sum_{cyc} a = 6 - \sum_{cyc} ab - \sum_{cyc} a =...
- 2,546
3
votes
Show that $ I = \int_0^1 \frac{dx}{\sqrt{\sin(2x) + \sin(x) + 3}} < \frac{1}{2} $?
We use the bound
$$\sqrt{\sin 2x + \sin x + 3} \ge \sqrt{3} + \frac{\sqrt{3}}{2}x - \frac{\sqrt 3}{4}x^2, \quad \forall x \in [0, 1].$$
The proof is given at the end.
(Note: The second order Taylor ...
- 36.4k
3
votes
Show that $ I = \int_0^1 \frac{dx}{\sqrt{\sin(2x) + \sin(x) + 3}} < \frac{1}{2} $?
Use the formula $\sin a + \sin b = 2 \sin({a + b \over 2})\cos({a - b \over 2})$ to rewrite your integral as
$$\int_0^1 {dx \over \sqrt{2\sin {3x \over 2}\cos {x \over 2} + 3}}$$
Here $\cos {x \over 2}...
- 44.4k
3
votes
Prove $3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$
We need to prove that:
$$(3a^3+4b^3+4c^3+abc)^2\geq4\left(\sqrt{b^3(4b^3+5a^3)}+\sqrt{c^3(4c^3+5a^3)}\right)^2$$ or $$9a^6+4(b^3+c^3)a^3+32b^3c^3+2abc(3a^3+4b^3+4c^3)+a^2b^2c^2\geq8\sqrt{b^3c^3(5a^3+...
- 197k
3
votes
Prove $3a+4(b+c)+\sqrt[3]{abc}\ge 2\left( \sqrt{b\left(4b+5a\right) }+ \sqrt{c\left(4c+5a\right) }\right).$
Sketch of a proof.
If $a = 0$, clearly the desired inequality is true.
If $a > 0$, WLOG, assume that $a = 1$.
The desired inequality is written as
$$\left(3 + 4(b + c) + \sqrt[3]{bc}\right)^2
\ge ...
- 36.4k
3
votes
Accepted
How to prove $\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$
Proof.
By using Cauchy-Schwarz and AM-GM inequality
\begin{align*}
\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}&=\sqrt{\frac{(a+b+c)(a^2(b+c)+abc)}{a^2bc}}\\&\ge \frac{a(...
- 1,874
3
votes
How to prove $\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 3+\frac{1}{3}\sum_{cyc}\left(\frac{b-c}{b+c}\right)^2.$
By the d8g3n1v9's beautiful idea we can prove a stronger inequality:
let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\sqrt{(a+b+c)\left(\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}\right)}\geq3+\...
- 197k
3
votes
Prove $\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge \sqrt{3x^3+6xyz}+\sqrt{3y^3+6xyz}+\sqrt{3z^3+6xyz}.$
As TATA box pointed out in the comments, OP's inequality is incorrect for when $ x = y \in (0.5, 2], z = \frac{ 2x-x^2 }{2x-1} $. EG See Wolfram Alpha's evaluation.
Proceeding from OP's inequality, ...
- 66.8k
2
votes
Prove $\frac{1}{\sqrt{2a+b+c}}+\frac{1}{\sqrt{2b+a+c}}+\frac{1}{\sqrt{2c+b+a}}\le \frac{9}{2}\cdot\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$
The Bacteria helps.
We need to prove that:
$$\sum_{cyc}\frac{1}{\sqrt{2a^2+b^2+c^2}}\leq\frac{9}{2(a+b+c)},$$ where $a$, $b$ and $c$ are positives.
Indeed, by C-S $$\sum_{cyc}\frac{1}{\sqrt{2a^2+b^2+c^...
- 197k
2
votes
Accepted
If $a^2+b^2+c^2=3,$ then prove $\sqrt{\frac{a+b}{4a+3}}+\sqrt{\frac{b+c}{4b+3}}+\sqrt{\frac{c+a}{4c+3}}\le 3\sqrt{\frac{2}{7}}.$
By C-S $$\sum_{cyc}\sqrt{\frac{a+b}{4a+3}}\leq\sqrt{\sum_{cyc}\frac{a+b}{(4a+3)(4a+8b+3c)}\sum_{cyc}(4a+8b+3c)}$$ and it's enough to prove that:
$$\sum_{cyc}\frac{a+b}{(4a+3)(4a+8b+3c)}\leq\frac{6}{35(...
- 197k
2
votes
Maximum value of $(1-a)(1-b)+(1-p)(1-q)$
As stated in the comments, you could check the method used by the AOPS community, but since you need help proceeding further in the trigonometric part, I will be posting that answer :
Since x, y are ...
- 475
2
votes
Finding $\small{\min\limits_{ab+bc+ca=1}\sqrt{a+2}+\sqrt{b+2}+\sqrt{c+2}- \sqrt{2-abc}.}$
Another way.
For $k=\sqrt{\frac{8}{3}}-1$ by Holder we obtain:
$$\left(\sum_{cyc}\sqrt{a+2}\right)^2\sum_{cyc}(a+2)^2(ka+b+c)^3\geq\left(\sum_{cyc}(a+2)(ka+b+c)\right)^3$$ and it's enough to prove ...
- 197k
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