IMO Problem Shortlist

For positive real numbers $x$, $y$, $z$, from the arithmetic-geometric-mean inequality, $$2x+y+z=(x+y)+(x+z)\ge 2\sqrt{(x+y)(x+z)}$$ we obtain $$\frac{1}{(2x+y+z{)}^2}\le \frac{1}{4(x+y)(x+z)}$$ Applying this to the left-hand side terms of the inequality to prove, we get $$\begin{array}{rl}\frac{1}{(2a+b+c{)}^2}& +\frac{1}{(2b+c+a{)}^2}+\frac{1}{(2c+a+b{)}^2}\\ & \le \frac{1}{4(a+b)(a+c)}+\frac{1}{4(b+c)(b+a)}+\frac{1}{4(c+a)(c+b)}\\ & =\frac{(b+c)+(c+a)+(a+b)}{4(a+b)(b+c)(c+a)}=\frac{a+b+c}{2(a+b)(b+c)(c+a)}\end{array}\text{\hspace{1em}(1)}$$ A second application of the inequality of the arithmetic-geometric mean yields $$a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b\ge 6abc$$ or, equivalently, $$\begin{array}{c}9(a+b)(b+c)(c+a)\ge 8(a+b+c)(ab+bc+ca)\end{array}\text{\hspace{1em}(2)}$$ The supposition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$ can be written as $$\begin{array}{c}ab+bc+ca=abc(a+b+c).\end{array}\text{\hspace{1em}(3)}$$ Applying the arithmetic-geometric-mean inequality $x^2{y}^2+x^2{z}^2\ge 2x^{2}yz$ thrice, we get $$a^2{b}^2+b^2{c}^2+c^2{a}^2\ge a^{2}bc+a{b}^{2}c+ab{c}^2$$ which is equivalent to $$\begin{array}{c}(ab+bc+ca{)}^2\ge 3abc(a+b+c)\end{array}\text{\hspace{1em}(4)}$$ Combining (1), (2), (3), and (4), we will finish the proof: $$\begin{array}{rl}\frac{a+b+c}{2(a+b)(b+c)(c+a)}& =\frac{(a+b+c)(ab+bc+ca)}{2(a+b)(b+c)(c+a)}\cdot \frac{ab+bc+ca}{abc(a+b+c)}\cdot \frac{abc(a+b+c)}{(ab+bc+ca{)}^2}\\ & \le \frac{9}{2\cdot 8}\cdot 1\cdot \frac{1}{3}=\frac{3}{16}\end{array}$$

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Alternative solution.

Equivalently, we prove the homogenized inequality $$\frac{(a+b+c{)}^2}{(2a+b+c{)}^2}+\frac{(a+b+c{)}^2}{(a+2b+c{)}^2}+\frac{(a+b+c{)}^2}{(a+b+2c{)}^2}\le \frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ for all positive real numbers $a$, $b$, $c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a$, $b$, $c>0$, fulfilling this condition, the inequality $$\begin{array}{c}\frac{1}{(1+a{)}^2}+\frac{1}{(1+b{)}^2}+\frac{1}{(1+c{)}^2}\le \frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\end{array}\text{\hspace{1em}(5)}$$ Applying Jensen's inequality to the function $f(x)=\frac{x}{(1+x{)}^2}$, which is concave for $0\le x\le 2$ and increasing for $0\le x\le 1$, we obtain $$\alpha \frac{a}{(1+a{)}^2}+\beta \frac{b}{(1+b{)}^2}+\gamma \frac{c}{(1+c{)}^2}\le (\alpha +\beta +\gamma )\frac{A}{(1+A{)}^2},\text{ where }A=\frac{\alpha a+\beta b+\gamma c}{\alpha +\beta +\gamma }.$$ Choosing $\alpha =\frac{1}{a}$, $\beta =\frac{1}{b}$, and $\gamma =\frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality $$A=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le \frac{a+b+c}{3}=\frac{1}{3}<1$$ Finally we prove (5): $$ $$\begin{array}{rl}\frac{1}{(1+a{)}^2}+\frac{1}{(1+b{)}^2}+\frac{1}{(1+c{)}^2}& \le \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{A}{(1+A{)}^2}\\ & \le \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{\frac{1}{3}}{{\left(1+\frac{1}{3}\right)}^2}=\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\end{array}$$