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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, $$\sum\limits_{cyc}\frac{a}{4a+4b+1}... 5 Given a,b,c,d>0 and a^2+b^2+c^2+d^2=1, prove$$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$$I was told that this is a pretty inequality to prove, but I have been unab... 0 Question: let a,b,c>0 show that$$(\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)$$Following it hard to pr... 6 How to prove this inequality$$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} ?$$Thanks 0 For a>0, b>0, c>0 and a^3+b^3+c^3=3 Prove that$$\frac{2ab}{\sqrt{c+3}}+\frac{2bc}{\sqrt{a+3}}+\frac{2ca}{\sqrt{b+3}}\le 3$$We have: abc\le \frac{a^3+b^3+c^3}{3}=1 \Leftrightarrow 2abc\left(\frac{1}{c\sqrt{c+3}}+\frac{1}{a\sqrt{a+3}}+\frac{1}{b\sqrt{b+3}}\right)\le 3 \Leftr... 2 For x>0, y>0, z>0 and x+y+z=3 find the minimize value of$$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)...