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Suppose that $x,y,z>0$ and $xyz=1$. Why does the inequality $$\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{x+z+1}\leq 1$$ holds? I couldn't see anything useful as I tried Jensen's inequality and calculus methods.

student
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1 Answers1

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Hint

If $a,b$ and $c$ be positive then $$\frac{1}{{{a}^{3}}+{{b}^{3}}+abc}+\frac{1}{{{b}^{3}}+{{c}^{3}}+abc}+\frac{1}{{{c}^{3}}+{{a}^{3}}+abc}\le \frac{1}{abc}$$ set $$a=\sqrt[3]{x}\quad,\quad b=\sqrt[3]{y}\quad,\quad c=\sqrt[3]{z}$$ Edit $$I=\frac{1}{{{a}^{3}}+{{b}^{3}}+abc}\le \frac{1}{a^2b+b^2a+abc}$$ $$J=\frac{1}{{{b}^{3}}+{{c}^{3}}+abc}\le \frac{1}{b^2c+c^2b+abc}$$ $$K=\frac{1}{{{c}^{3}}+{{a}^{3}}+abc}\le \frac{1}{c^2a+a^2c+abc}$$ we have $$I+J+K\le\frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$ $$I+J+K\le\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$ $$I+J+K\le\frac{1}{abc}$$

Behrouz Maleki
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  • I'm not really sure about this. It looks like you just set $a^3=x,b^3=y,c^3=z$ and watched how it modifies the inequality. Is the inequality you got a well known? It is new for me and I don't know how to prove it. – student Jun 30 '16 at 11:54
  • Your Inequality is not homogeneous but mine is a homogeneous inequality. – Behrouz Maleki Jun 30 '16 at 11:58
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    This is a first step. Now to the proof... – Did Jun 30 '16 at 12:06