An inequality

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An inequality

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#1 h Mar 20, 2025, 10:28 AM

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Let a,b,c>=0: ab+bc+ca=3 then maximize P=\frac{a^2b+b^2c+c^2a+9}{a+b+c}+\frac{abc}{2}.

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sqing

#2 h Mar 20, 2025, 1:14 PM

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Let $a,b,c\geq 0$ and $ab+bc+ca=3.$ Prove that $\frac{a^2b+b^2c+c^2a+9}{a+b+c}+\frac{abc}{2}\leq \frac{9}{2}$
Good.

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#5 h Mar 20, 2025, 2:47 PM

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sqing, can you help me?

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#6 h Yesterday at 4:53 AM • 1 Y

Y by Quantum-Phantom

WLOG $b$ be the median of $\{a, b, c\}$ . Then
$(9-2P)(a+b+c) = (2-b)(a+b+c-3)ac + 3(a+c)(b-1)^2 + 2c(a-b)(b-c) \geq 0.$ Equality holds when $(a, b, c) = (1, 1, 1)$ or $(a, b, c) = (3, 1, 0)$ .

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Quantum-Phantom

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Quantum-Phantom

#7 h Yesterday at 9:52 AM • 1 Y

Y by MS_asdfgzxcvb

@lbh_qys, how did you get this decomposition?

Alternative proof: WLOG let $x\ge y\ge z$ be an arrangement of $a$ , $b$ , $c$ , and set $x+z=2p$ , $xz-q^2$ . Hence,
$0\ge(x-y)(z-y)=q^2-2py+y^2,~3=xy+yz+zx=2py+q^2.$ Combining the two relations we get $3-4py+y^2\le0$ , so $p\ge\tfrac{3+y^2}{4y}$ . By rearrangement inequality,
$\sum_{\rm cyc}a^2b=\sum_{\rm cyc}ab\cdot a\le xy\cdot x+xz\cdot y+yz\cdot z=\left(4p^2-q^2\right)y.$ Thus, it suffices to show that
$\frac{\left(4p^2-q^2\right)y+9}{2p+y}+\frac{q^2y}2\le\frac92,$ or
$-18+18 p+9 y-8 p^2 y+q^2 \left(2 y-2 p y-y^2\right)\ge0,$ or by replacing $q^2$ with $3-2py$ ,
$4 p^2 (-2+y) y-3 \left(6-5 y+y^2\right)+2 p \left(9-3 y-2 y^2+y^3\right) \ge0.$ Clearly $y\le2$ , so we need to prove that a concave function of $p$ is non-negative. It suffices to check boarder cases, i.e. for $p=\tfrac{3+y^2}{4y}$ and $p=\tfrac3{2y}$ . These two cases respectively reduces to
$\frac3{4y}\left(y^3-3y+12\right)(y-1)^2\ge0,~\frac9y(y-1)^2\ge0.$ We are done.

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