Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

64. If a,b,ca, b, c are positive numbers, then
a+b+c+a2b+b2c+c2a⩾6(a2+b2+c2)a+b+ca+b+c+\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geqslant \frac{6\left(a^{2}+b^{2}+c^{2}\right)}{a+b+c}

Solution

64. (2007.03.10) Proof: Since
a2b+b2c+c2a=∑[−b+2a+(a−b)2b]=∑a+∑(a−b)2b\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}=\sum\left[-b+2 a+\frac{(a-b)^{2}}{b}\right]=\sum a+\sum \frac{(a-b)^{2}}{b}

Therefore,
∑a⋅(2∑a+∑a2b)=2(∑a)2+∑a⋅∑(a−b)2b⩾2(∑a)2+(∑∣a−b∣)2⩾2(∑a)2+4(∑a2−∑bc)=6∑a2\begin{array}{l} \sum a \cdot\left(2 \sum a+\sum \frac{a^{2}}{b}\right)=2\left(\sum a\right)^{2}+\sum a \cdot \sum \frac{(a-b)^{2}}{b} \geqslant \\ 2\left(\sum a\right)^{2}+\left(\sum|a-b|\right)^{2} \geqslant 2\left(\sum a\right)^{2}+4\left(\sum a^{2}-\sum b c\right)= \\ 6 \sum a^{2} \end{array}

Thus, the original inequality is proved.

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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.