$\sum \frac{a}{b+c+d}\le \frac{2\sum a^2}{\sum ab}$ if $\sum a =4$

Question

Let $a, b, c, d$ positive real numbers such that $a+b+c+d=4$. Prove that $$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\le \frac{2(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}.$$

My idea is to cancel the denominators, expand the inequality and use Muirhead, but the calculations are terrible... So I'm looking for a smarter solution.

Answer 1

Fill in any missing gaps.

• Multiplying by $(ab+ac+ad+bc+bd+cd)$, show that the inequality is equivalent to $$\sum \frac{ abc+abd+acd}{b+c+d} \leq \sum a^2. $$
• Show that $$ \frac{ abc+abd+acd}{b+c+d} \leq \frac{ab+ac+ad}{3} \leq \frac{3a^2+b^2+c^2+d^2}{6}$$
• Hence, summing up the cyclic inequalities, the conclusion follows.

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Note: This could be combined into 1 (non-obvious) step by showing that

$$ \frac{a}{b+c+d} \leq \frac{ 3a^2 + b^2+c^2+d^2} { 3(ab+ac+ad+bc+bd+cd)}.$$