Need help on this 2

(a+b)(b+c)(c+a)abc343(a2a2+bc+b2b2+ac+c2c2+ab)\large \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq\frac{4}{3}\bigg(\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\bigg) If a,ba,b and cc are positive reals, prove the inequality above.

I've proven (a+b)(b+c)(c+a)abc32\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 2 So now I have to prove a2a2+bc+b2b2+ac+c2c2+ab32\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\leq\frac{3}{2} I've tried to prove it but then realised that what I was trying to prove is wrong, can somebody help me?

Note by P C
3 years, 7 months ago

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How did you proved the first part

Department 8 - 3 years, 7 months ago

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First can be proved easily using AM-GM inequality.

Harsh Shrivastava - 3 years, 7 months ago

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Yes just saw that

Department 8 - 3 years, 7 months ago

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I proved it up to the lhs is greater than or equal to 427abc/( (a+b+c)^3+27abc)

Department 8 - 3 years, 7 months ago

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can you show me how?

P C - 3 years, 7 months ago

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If you are on Slack, I can send you there.

Department 8 - 3 years, 7 months ago

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@Department 8 yes I am, can you send it to me?

P C - 3 years, 7 months ago

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@P C I am busy with my finals so I would send you it later make sure you check it

Department 8 - 3 years, 7 months ago

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@Department 8 alright, good luck dude

P C - 3 years, 7 months ago

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@P C Thanks, but please check it out later, but it may contain flaws and feel free to spot the mistakes

Department 8 - 3 years, 7 months ago

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