18th Turkish Mathematical Olympiad

We first observe that $x^4+3\ge (x+1{)}^2$ as $x^4+3-(x+1{)}^2=(x-1{)}^2(x^2+2x+2)\ge 0$. Therefore it suffices to prove that $$\sum _{i=1}^n\frac{a_i}{a_i+1}\le \frac{1}{2}\sum _{i=1}^n\frac{1}{a_i}$$ for all positive integers $n$ and positive real numbers $a_1,a_2,\dots ,a_n$ satisfying the condition $a_1{a}_2\cdots a_n=1$.

Next we observe that $$\frac{1}{2xy}-\frac{xy}{xy+1}\le \frac{1}{2x}+\frac{1}{2y}-\frac{x}{x+1}-\frac{y}{y+1}(\ast )$$ for $0<x\le 1\le y$. This can be seen using the fact that $$0\le (1-x)(y-1)(2x^2{y}^2+x^{2}y+x{y}^2+xy+x+y+1)$$ after collecting all the terms to the right side of the inequality and getting a common denominator.

Let $f_k(x_1,x_2,\dots ,x_k)=\frac{1}{2}{\sum }_{i=1}^k\frac{1}{x_i}-{\sum }_{i=1}^k\frac{x_i}{x_i+1}$. We will prove by induction on $k$ that $x_1{x}_2\cdots x_k=1⟹f_k(x_1,x_2,\dots ,x_k)\ge 0$.

For $k=1$, $x_1=1$ and $f_1(x_1)=0$.

Suppose $k>1$ and $x_1{x}_2\cdots x_k=1$. If $x_1\le x_2\le \cdots \le x_k$, then $x_1\le 1\le x_k$, and using (*) we get $f_k(x_1,x_2,\dots ,x_k)\ge f_{k-1}(x_1{x}_k,x_2,\dots ,x_{k-1})\ge 0$.