Answered: "Proof of one inequality a+b+c≤a3bc+b3ca+c3ab"

Pamela Meyer

Answered question

2022-01-07

Proof of one inequality

a + b + c \leq \frac{a^{3}}{b c} + \frac{b^{3}}{c a} + \frac{c^{3}}{a b}

Answer & Explanation

Neil Dismukes

Beginner2022-01-08Added 37 answers

Using Cauchy-Schwarz Inequality twice:

a^{4} + b^{4} + c^{4} \geq a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq a b^{2} c + b a^{2} c + a c^{2} b = a b c (a + b + c)

Dawn Neal

Beginner2022-01-09Added 35 answers

I have come up with an answer with myself. Using CS inequality

(a^{4} + b^{4} + c^{4}) (1 + 1 + 1) \geq {(a^{2} + b^{2} + c^{2})}^{2}

(a^{2} + b^{2} + c^{2}) (1 + 1 + 1) \geq {(a + b + c)}^{2}

Hence we have

a^{4} + b^{4} + c^{4} \geq \frac{{(a + b + c)}^{4}}{27} = (a + b + c) {(\frac{a + b + c}{3})}^{3} \geq a b c (a + b + c)

Vasquez

Expert2022-01-11Added 669 answers

By Holder
$\sum_{cyc} \frac{a^{3}}{b c} \geq \frac{(a + b + c)^{3}}{3 (a b + a c + b c)} = \frac{(a + b + c) \cdot (a + b + c)^{2}}{3 (a b + a c + b c)} \geq a + b + c$

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