Proof of one inequality a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}

Pamela Meyer

Pamela Meyer

Answered question

2022-01-07

Proof of one inequality
a+b+ca3bc+b3ca+c3ab

Answer & Explanation

Neil Dismukes

Neil Dismukes

Beginner2022-01-08Added 37 answers

Using Cauchy-Schwarz Inequality twice:
a4+b4+c4a2b2+b2c2+c2a2ab2c+ba2c+ac2b=abc(a+b+c)
Dawn Neal

Dawn Neal

Beginner2022-01-09Added 35 answers

I have come up with an answer with myself. Using CS inequality
(a4+b4+c4)(1+1+1)(a2+b2+c2)2
(a2+b2+c2)(1+1+1)(a+b+c)2
Hence we have
a4+b4+c4(a+b+c)427=(a+b+c)(a+b+c3)3abc(a+b+c)
Vasquez

Vasquez

Expert2022-01-11Added 669 answers

By Holder
cyca3bc(a+b+c)33(ab+ac+bc)=(a+b+c)(a+b+c)23(ab+ac+bc)a+b+c

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