Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that: $$\frac{a}{4a+4b+1}+\frac{b}{4b+4c+1}+\frac{c}{4c+4a+1}\leq\frac{1}{3}$$ My trying. $$3=a^2+b^2+c^2\geq\frac{1}{3}(a+b+c)^2,$$ which gives $1\geq\frac{a+b+c}{3}$. Thus, $$\sum\limits_{cyc}\frac{a}{4a+4b+1}...

Given $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1$, prove $$a+b+c+d\ge a^3+b^3+c^3+d^3+ab+ac+ad+bc+bd+cd$$ The inequality can be written in the condensed form $$\sum\limits_{Sym}a\ge\sum\limits_{Sym}a^3+\sum\limits_{Sym}ab$$ I was told that this is a pretty inequality to prove, but I have been unab...

Question: let $a,b,c>0$ show that $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a+b+c)^3\ge 27(ab+bc+ac)^2$$ since $$(a+b+c)^2\ge 3(ab+bc+ac)\Longleftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\ge 0$$ it is enough to show that $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2(a+b+c)\ge 9(ab+bc+ac)$$ Following it hard to pr...

How to prove this inequality $$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} ?$$ Thanks

For $a>0$, $b>0$, $c>0$ and $a^3+b^3+c^3=3$ Prove that $$\frac{2ab}{\sqrt{c+3}}+\frac{2bc}{\sqrt{a+3}}+\frac{2ca}{\sqrt{b+3}}\le 3$$ We have: $abc\le \frac{a^3+b^3+c^3}{3}=1$ $\Leftrightarrow 2abc\left(\frac{1}{c\sqrt{c+3}}+\frac{1}{a\sqrt{a+3}}+\frac{1}{b\sqrt{b+3}}\right)\le 3$ $\Leftr...

For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$ We have: $P=\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)...