Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.

Prove the inequality. $a^2+b^2+1≥ab+a+b$ I try so many methods, But I have not been successful in any way.Because, I can not find "hint".

$x,y,z >0$, prove $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$ Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few yea...

$x,y,z >0$ and $x+y+z=3$, prove $$\tag{1}\frac{xy}{5y^3+4}+\frac{yz}{5z^3+4}+\frac{zx}{5x^3+4} \leqslant \frac13$$ My first attempt is to use Jensen's inequality. Hence I consider the function $$f(x) =\frac{x}{5x^3+4}$$ Compute second derivative we have $$\tag{2}f''(x)=\frac{30x^2(5x^3-8)}...

Let $ab+bc+ca=1$. Prove that $2 \ge \sqrt{1+a^2} + \sqrt{1+b^2}+\sqrt{1+c^2}-a-b-c \geq \sqrt3 $.

I have inequality $$x\cdot\ln(x) + y\cdot\ln(y) \geq (x+y)\cdot \ln(x+y)$$ I transformated it to $$2^{x+y}\cdot x^x\cdot y^y\geq(x+y)^{x+y}$$ And I got stuck. Please help

In $\Delta ABC$,let $x=\sin{A},y=\sin{B},z=\sin{C}$,show that $$(x+y+z)^2\ge 4(x^2y^2+y^2z^2+z^2x^2)$$ I tried C-S and more, but without success. I am looking for an human proof, which we can use during competition.

Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$. I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.

For real numbers $a,b,c \in [0,1]$ prove inequality $$\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}+abc\le \frac52$$ I tried AM-GM, Buffalo way. I do not know how to solve this problem

Let $a, b, c, d \in \mathbb{R}_{>0}$, then prove that $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$ Can this be done without using AM-GM inequality, or without using any identity/theorem of inequality? I don't want it to be concise or elegant, I just want rigorous steps that sh...

If $a,b,c,d$ are positive real number, Then prove that $$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2}>a+b+c+d$$ $\bf{Attempt:}$ $$\frac{abcd}{a^3}+\frac{abcd}{b^3}+\frac{abcd}{c^3}+\frac{abcd}{d^4}$$ Using Arithmetic Geometric Inequality$$abcd\bigg[a^{-3}+b^{-3}+c^{-3}+d^...

give the postive intger $n\ge 2$,and postive real numbers $a<b$ if the real numbers such $x_{1},x_{2},\cdots,x_{n}\in[a,b]$ find the maximum of the value $$\dfrac{\frac{x^2_{1}}{x_{2}}+\frac{x^2_{2}}{x_{3}}+\cdots+\frac{x^2_{n-1}}{x_{n}}+\frac{x^2_{n}}{x_{1}}}{x_{1}+x_{2}+\cdots+x_{n}}$$ it seem...

$a,b,c >0$ and $abc=1$, prove $$\frac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geqslant \frac{3}{\sqrt[4]{2}}$$ 1. I tried rearrangement and AM-GM but fail. 2. I think the power of $\frac14$ is tough. I can prove the easier inequality $$\frac{1}{a^3...

How can one prove/disprove that $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$ where $R$ and $r$ denote the usual circum and inradii respectively. I know that $R=\frac{abc}{4\Delta}$ and $r=\frac{\Delta}{s}$, where $\Delta$ denotes area of triangle, and $s$ the semi perimeter. Any ideas. Thanks bef...

$a, b, c$ are positive real numbers such that $ab+bc+ca=3abc$ Prove∶ $$\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\;\;\leq\; 3$$

If $abc=1$ then $$\frac1{a+b+1}+\frac1{b+c+1}+\frac1{c+a+1}\le1$$ I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?

Let $x$, $y$, $z$ be positive. Prove the inequality $$4(x+y+z)^3 \ge 27(yx^2+zy^2+xz^2+xyz)$$ I have no idea of how the proof should look like, tried to get rid of the braces but it seems to be the wrong way, as the equation becomes very long and has very different positive and negative coeffici...

Let $a$, $b$, and $c$ be posistive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that: $$ ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc. $$ Let us consider the following proofs. $$ a^{2}+b^{2}+c^{2} \geq ab+bc+ca $$ By the Arit...

Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$ Attempt I tried using HM-GM...

Question: let $x,y,z>0$ and such $xyz=1$, show that $$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$ My idea: use AM-GM inequality $$x^3+x^3+1\ge 3x^2$$ $$y^3+y^3+1\ge 3y^2$$ $$z^3+z^3+1\ge 3z^2$$ so $$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$ But this is not my inequality,so How prove it? I know this co...

Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$($a,b,c$ are positive real numbers). There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!

Let $x$, $y$ and $z$ be positive reals with $xy+yz+zx=1$. Prove the inequality $$xyz(x+y)(y+z)(z+x)\ge (1-x^2)(1-y^2)(1-z^2).$$