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# Inequality 22(Nguyen Van Thach)

Problem:

Let $\displaystyle a,b,c,d$ be positive real numbers. Prove that

\displaystyle \begin{aligned}\frac{abc}{(d+a)(d+b)(d+c)}+\frac{bcd}{(a+b)(a+c)(a+d)}&+\frac{cda}{(b+c)(b+d)(b+a)}+\frac{dab}{(c+a)(c+b)(c+d)}\\&\geq\frac{1}{2}\end{aligned}.

Solution:

Denote by $\displaystyle x,y,z,t$ the $\displaystyle \frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ respectively. Then the inequality takes the form:

\displaystyle \begin{aligned}\frac{x^3}{(x+y)(x+z)(z+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}&+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\\&\geq \frac{1}{2}\end{aligned}.

Assume, without loss of generality, that $\displaystyle x+y+z+t=4$. Then from the AM-GM inequality we get that

$\displaystyle \sum_{cyc}\frac{x^3}{(x+y)(x+z)(x+t)}\geq \sum_{cyc}\frac{x^3}{\left(x+\frac{y+z+t}{3}\right)^3}=\frac{27}{8}\cdot \sum_{cyc}\frac{x^3}{(x+2)^3}$.

So, it is enough to prove that

$\displaystyle \sum_{cyc}\frac{x^3}{(x+2)^3}\geq \frac{4}{27}$.

But the last relation can be rewritten as $\displaystyle \sum_{cyc}\frac{x^3}{(x+2)^3}\geq \sum_{cyc}\frac{2x-1}{27}$.

Bringing everything in the left hand side the given inequality is of the form

$\displaystyle \frac{2}{27}\cdot\sum_{cyc}(-x^2+6x+4)(1-x)^2\geq 0$, which is obviously true, Q.E.D.