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While trying to prove it, I proved the following two inequalities:

$a^4+b^4+c^4\ge abc(a+b+c)$ and

$(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$

The later one, on some simplification gives

$a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$

But we can't claim that $ab+bc+ca\ge a+b+c$ for all positive $a, b, c.$ So this doesn't help. So am not quite sure how to approach the inequality in question. Please suggest.. Thanks in advance. (BTW can we use Cauchy-Schwarz's inequality? I tried but couldn't think of a proper choice for two triplets.)

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Using “Titu's Lemma” (also called “Engel's form” of the Cauchy-Schwarz inequality) you have for $a, b, c > 0$ $$ \frac{a^3b + b^3c + c^3a}{abc} = \frac{a^2}{c} + \frac{b^2}{a} +\frac{c^2}{b} \ge \frac{(a+b+c)^2}{c+a+b} = a+b+c $$

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We need to prove that $$\sum_{cyc}\frac{a^2}{c}\geq\sum_{cyc}a,$$ which is true by Rearrangement: $$\sum_{cyc}\frac{a^2}{c}=\sum_{cyc}a^2\cdot\frac{1}{c}\geq \sum_{cyc}a^2\cdot\frac{1}{a}=\sum_{cyc}a.$$

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Another way.

By AM-GM $$\sum_{cyc}\left(\frac{a^2}{c}-a\right)=\sum_{cyc}\left(\frac{a^2}{c}+c-2a\right)\geq\sum_{cyc}\left(2\sqrt{\frac{a^2}{c}\cdot c}-2a\right)=0.$$

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  • $\begingroup$ Got it.. (A small edit in the second last line: the second c should be inside the square root). Thanks again. $\endgroup$ – Dhrubajyoti Bhattacharjee Jun 3 '20 at 19:37
  • $\begingroup$ I have taken the liberty to fix that typo. $\endgroup$ – Martin R Jun 3 '20 at 19:43
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Suppose $c=\min\{a,b,c\}.$ We have $$\begin{aligned}a^3b+b^3c+c^3a-abc(a+b+c)&=c(a^3+b^3-a^2b-ab^2)+a(c^3+a^2b-bc^2-ca^2)\\&=c(a+b)(a-b)^2+a(c+a)(a-c)(b-c) \geqslant 0.\end{aligned}$$

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    $\begingroup$ You mean $a^3 b+b^3 c+c^3 a -abc(a+b+c)$ $\endgroup$ – tthnew Jun 4 '20 at 9:11
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We have$:$ $$a^3 b+b^3 c+c^3 a-abc(a+b+c)$$ $$=bc \,\left( a-b \right) ^{2}+ac \left( b-c \right) ^{2}+ab \left( c-a \right) ^{2} \geqq 0$$ Done.

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