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#1 h Jan 29, 2025, 10:27 AM • 1 Y

Y by PikaPika999

Given $a,b,c$ are positive real numbers satisfied $a^3+b^3+c^3=3$ . Prove that:
$\sqrt{2ab+5c^{2}+2a}+\sqrt{2bc+5a^{2}+2b}+\sqrt{2ac+5b^{2}+2c}\le3\sqrt{3\left(a+b+c\right)}$

This post has been edited 1 time. Last edited by nikiiiita, Jan 29, 2025, 10:29 AM

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arqady

#3 h Jan 30, 2025, 8:34 AM • 2 Y

Y by nikiiiita, PikaPika999

nikiiiita wrote:

Given $a,b,c$ are positive real numbers satisfied $a^3+b^3+c^3=3$ . Prove that:
$\sqrt{2ab+5c^{2}+2a}+\sqrt{2bc+5a^{2}+2b}+\sqrt{2ac+5b^{2}+2c}\le3\sqrt{3\left(a+b+c\right)}$

By C-S $\sum_{cyc}\sqrt{5a^2+2bc+2b}\leq\sqrt{\sum_{cyc}\frac{5a^2+2bc+2b}{2a+b+c}\sum_{cyc}(2a+b+c)}$ and it's enough to prove that:
$\sum_{cyc}\frac{5a^2+2bc+2b}{2a+b+c}\leq\frac{27}{4}$ or
$\sum_{cyc}(38a^3+141a^2b+149a^2c+120abc)\geq\sum_{cyc}(20a^4+76a^3b+76a^3c+120a^2b^2+156a^2bc)$ and after using AM-GM it's enough to prove $\sum_{cyc}(38a^3+141a^2b+141a^2c+128abc)\geq\sum_{cyc}(20a^4+76a^3b+76a^3c+120a^2b^2+156a^2bc),$ which is smooth.

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#4 h Jan 30, 2025, 12:50 PM • 1 Y

Y by PikaPika999

Nice, thanks
For $a,b,c$ are non-negative reals, $a+b+c+abc=4$ . Prove that:
$\sqrt{ab+c^{2}+a}+\sqrt{bc+a^{2}+b}+\sqrt{ac+b^{2}+c} \ge 3\sqrt{\left(a+b+c\right)}$ Equality sign occurs for: $a=b=c=1$ or $a=b=0,c=4$ and its permutations
Is it old?

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30219 posts

arqady

#5 h Jan 30, 2025, 4:27 PM • 1 Y

Y by PikaPika999

nikiiiita wrote:

For $a,b,c$ are non-negative reals, $a+b+c+abc=4$ . Prove that:
$\sqrt{ab+c^{2}+a}+\sqrt{bc+a^{2}+b}+\sqrt{ac+b^{2}+c} \ge 3\sqrt{\left(a+b+c\right)}$ Equality sign occurs for: $a=b=c=1$ or $a=b=0,c=4$ and its permutations
Is it old?

It's new for me. I think, it's very difficult to find a nice solution.

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#6 h Feb 13, 2025, 12:32 PM • 1 Y

Y by PikaPika999

bump for #4

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#7 h Apr 6, 2025, 8:58 PM • 1 Y

Y by PikaPika999

Any ideas?

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