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Let $a,b,c>0$. Prove that: $$a b \sqrt{a b}+b c \sqrt{b c}+c a \sqrt{c a} \leqslant a b c+\frac{1}{2} \sqrt[3]{\frac{\left(a^{2}+b c\right)^{2}\left(b^{2}+c a\right)^{2}\left(c^{2}+a b\right)^{2}}{a b c}}$$

I really don't have many ideas in this problem. First I thought of using AM-GM: $a^2+bc\ge 2a\sqrt{bc}$ and so on, or using Holder $(\dfrac{a^3}{bc}+2a+\dfrac{bc}{a})(\dfrac{b^3}{ac}+2b+\dfrac{ca}{b})(\dfrac{c^3}{ab}+2c+\dfrac{ab}{c })$ but all make the inequality into the form $x^3+y^3+z^3\le3xyz$, which is of course incorrect. I thought about using derivatives but they don't work either

Can anyone give me a different way of thinking?

trungbk
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1 Answers1

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Using AM-GM, it suffices to prove that $$ab\frac{a + b}{2} + bc\frac{b + c}{2} + ca\frac{c + a}{2} \le abc + \frac12\sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}$$ or $$a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2 - 2abc \le \sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}.$$

Since the inequality is homogeneous, WLOG, assume that $abc = 1$.

Let $p = a + b + c, q = ab + bc + ca, r = abc = 1$. We have $p, q \ge 3$.

The inequality is written as $$pq - 5r \le \sqrt[3]{\frac{(p^3r + q^3 - 6pqr + 8r^2)^2}{r}}$$ or $$pq - 5 \le \sqrt[3]{(p^3 + q^3 - 6pq + 8)^2}.$$

Using AM-GM, we have $$p^3 + q^3 = (p + q)(p^2 - pq + q^2) \ge (p + q)pq \ge 2\sqrt{pq}\, pq.$$ Using $p, q \ge 3$, we have $$2\sqrt{pq}\, pq \ge 6pq.$$

It suffices to prove that $$pq - 5 \le \sqrt[3]{(2\sqrt{pq}\,pq - 6pq + 8)^2}.$$ Letting $x = \sqrt{pq} \ge 3$, the inequality becomes $$x^2 - 5 \le \sqrt[3]{(2x^3 - 6x^2 + 8)^2}$$ or $$(x^2 - 5)^3 \le (2x^3 - 6x^2 + 8)^2$$ or $$(3x^4 - 6x^3 - 12x^2 + 14x + 21)(x - 3)^2 \ge 0$$ which is true.

We are done.

River Li
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