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Let $a,b,c>0$, prove that

$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2.$$ I tried using the equality $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ and the Schur inequality but it's not very helpful.

Thanks.

zarathustra
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user169301
  • 283

3 Answers3

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WOLG:$a\ge b\ge c$ we have $$2b(a+c)^2-(a+b)(b+c)(a+c)=(a+c)(a-b)(b-c)\ge 0$$ so $$\dfrac{8abc}{(a+b)(b+c)(a+c)}\ge\dfrac{4ac}{(a+c)^2}$$ so we only prove $$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$ since $$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}- 2=\dfrac{(a^2+c^2 -ab-bc)^2}{(a+c)^2(ab+bc+ac)}\ge 0$$

math110
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3

I would like to use @math 110 idea with a little difference when we prove that

$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{4ac}{(a+c)^2}\ge 2$$

Since $\dfrac{a^2+b^2+c^2}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}-2$, we can rewrite the inequality as

$\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 4-\dfrac{4ac}{(a+c)^2}=\dfrac{4(a^2+ac+c^2)}{(a+c)^2}$

Or $4(a^2+ac+c^2)(ab+bc+ac)\le (a+c)^2(a+b+c)^2$

Using AM-GM, we have $4(a^2+ac+c^2)(ab+bc+ac)\le (a^2+ac+c^2+ab+bc+ac)^2 =[(a+c)^2+b(a+c)]^2 =(a+c)^2(a+b+c)^2$

user169478
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We need to prove that $$\frac{a^2+b^2+c^2}{ab+ac+bc}-1\geq1-\frac{8abc}{(a+b)(a+c)(b+c)}$$ or $$\sum\limits_{cyc}\frac{(a-b)^2}{2(ab+ac+bc)}\geq\frac{\sum\limits_{cyc}c(a-b)^2}{(a+b)(a+c)(b+c)}$$ or $$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq0.$$ Let $a\geq b\geq c$. Hence, $(a-c)^2\geq (b-c)^2$, $$a^2b+a^2c+b^2a+b^2c-c^2a-c^2b\geq0$$ and $$a^2b+a^2c+c^2a+c^2b-b^2a-b^2c\geq0.$$ Thus, $$\sum_{cyc}(a-b)^2(a^2b+a^2c+b^2a+b^2c-c^2a-c^2b)\geq$$ $$\geq(b-c)^2(a^2b+a^2c+c^2a+c^2b-b^2a-b^2c+b^2a+b^2a+c^2a+c^2b-a^2b-a^2c)=$$ $$=2(b-c)^2c^2(a+b)\geq0.$$ Done!

Michael Rozenberg
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