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Algebra Difficulty 6.2 National olympiad Prove it

Let 3.21a,b,c,d3.21 a, b, c, d be non-negative real numbers. Prove
1+7ab+c+d+1+7bc+d+a+1+7cd+a+b+1+7da+b+c4103\begin{array}{c} \sqrt{1+\frac{7 a}{b+c+d}}+\sqrt{1+\frac{7 b}{c+d+a}}+\sqrt{1+\frac{7 c}{d+a+b}}+ \\ \sqrt{1+\frac{7 d}{a+b+c}} \geqslant 4 \sqrt{\frac{10}{3}} \end{array}

Solution

By the homogeneity of the inequality, without loss of generality, assume a+b+c+d=4a+b+c+d=4. Let f(x)=1+7x4xf(x)=\sqrt{1+\frac{7 x}{4-x}}, then we have
f(x)13215(8+7x)f(x) \geqslant \frac{1}{3} \sqrt{\frac{2}{15}}(8+7 x)

The above inequality holds because
14(x1)2(2+7x)135(4x)0\frac{14(x-1)^{2}(2+7 x)}{135(4-x)} \geqslant 0

Substituting a,b,c,da, b, c, d into f(x)f(x) respectively and adding them up yields the desired inequality.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.