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Inequality 32(Vasile Cirtoaje)

The following inequality and its solution is dedicated to my beloved teacher, Christos Patilas…


If \displaystyle a,b,c,d are positive real numbers, with \displaystyle abcd=1, prove that

\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}+\frac{1}{(1+d)(1+d^2)}\geq 1.


Without loss of generality assume that \displaystyle a\geq b\geq c\geq d. Then from Chebyshev’s inequality we have

\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}\geq \frac{1}{3}\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right).

Lemma (Vasile Cirtoaje):

If \displaystyle a\geq b\geq c\geq d and \displaystyle abcd=1 then it holds that

\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{3}{1+\sqrt[3]{abc}}.

Proof of the Lemma:

We know that \displaystyle \frac{1}{1+a}+\frac{1}{1+b}\geq \frac{2}{1+\sqrt{ab}}. So, it suffices to prove that \displaystyle \frac{1}{1+c}+\frac{2}{1+\sqrt{ab}}\geq \frac{3}{1+\sqrt[3]{abc}}.

Let us denote by \displaystyle x=\sqrt{ab},y=\sqrt[3]{abc}\Longrightarrow c=\frac{y^3}{x^2}. Substituting them to the above inequality we get that

\displaystyle \frac{1}{1+c}+\frac{2}{1+\sqrt{ab}}-\frac{3}{1+\sqrt[3]{abc}}=\frac{x^2}{x^2+y^3}+\frac{2}{1+x}-\frac{3}{1+y},

which reduces to the inequality

\displaystyle \frac{(x-y)^2\left[2y^2-y+x(y-2)\right]}{(1+x)(1+y)(x^2+y^3)},

which is obvious since \displaystyle 2y^2-y+(y-2)x\geq 2y^2-y+(y-2)y^3=y(y-1)(y^2-y+1)\geq 0.

Back to our inequality now, from the above lemma we deduce that:

\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{3}{1+\sqrt[3]{abc}}\wedge \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\geq \frac{3}{1+\sqrt[3]{a^2b^2c^2}}.

For convenience denote by \displaystyle k the \displaystyle \sqrt[3]{abc}. Therefore we have that

\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}\geq \frac{3}{(1+k)(1+k^2)}.

Thus it remains to prove that

\displaystyle \frac{3}{(1+k)(1+k^2)}+\frac{1}{(1+d)(1+d^2)}\geq 1.

But \displaystyle abcd=1. So, the last fraction is of the form \displaystyle \frac{1}{\left(1+\frac{1}{k^3}\right)\left(1+\frac{1}{k^6}\right)}.

After that we get

\displaystyle \frac{1}{\left(1+\frac{1}{k^3}\right)\left(1+\frac{1}{k^6}\right)}+\frac{3}{(1+k)(1+k^2)}\geq 1.

Conclusion follows from the obvious inequality  \displaystyle \frac{(k-1)^2(2k^4+k^3+k+2)}{(k^3+1)(k^6+1)}\geq 0


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