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Algebra Difficulty 6.9 National olympiad Prove it

Example 6 (31st IMO Preliminary Question) Let a,b,ca, b, c be positive real numbers. Prove that:
(a2+ab+b2)(b2+bc+c2)(c2+ca+a2)≥(ab+bc+ca)3\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geq(a b+b c+c a)^{3}

Solution

Proof: It is easy to prove:
a2+ab+b2≥34(a+b)2b2+bc+c2≥34(b+c)2c2+ca+a2≥34(c+a)2\begin{array}{l} a^{2}+a b+b^{2} \geq \frac{3}{4}(a+b)^{2} \\ b^{2}+b c+c^{2} \geq \frac{3}{4}(b+c)^{2} \\ c^{2}+c a+a^{2} \geq \frac{3}{4}(c+a)^{2} \end{array}

Therefore, to prove the original inequality, it suffices to prove
27(a+b)2(b+c)2(c+a)2≥64(ab+bc+ca)327(a+b)^{2}(b+c)^{2}(c+a)^{2} \geq 64(a b+b c+c a)^{3}

Since (abc)2=ab⋅bc⋅ca≤(ab+bc+ca3)3(a b c)^{2}=a b \cdot b c \cdot c a \leq\left(\frac{a b+b c+c a}{3}\right)^{3},
we have abc≤(ab+bc+ca3)32a b c \leq\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}},
(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc≥3(ab+bc+ca)⋅(ab+bc+ca)−abc≥3(ab+bc+ca)32−(ab+bc+ca3)32=893(ab+bc+ca)32\begin{aligned} (a+b)(b+c)(c & +a)=(a+b+c)(a b+b c+c a)-a b c \\ & \geq \sqrt{3(a b+b c+c a)} \cdot(a b+b c+c a)-a b c \\ & \geq \sqrt{3}(a b+b c+c a)^{\frac{3}{2}}-\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}} \\ & =\frac{8}{9} \sqrt{3}(a b+b c+c a)^{\frac{3}{2}} \end{aligned}

Thus, 27(a+b)2(b+c)2(c+a)2≥64(ab+bc+ca)327(a+b)^{2}(b+c)^{2}(c+a)^{2} \geq 64(a b+b c+c a)^{3}.
Hence, the 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.