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pennypc123456789

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pennypc123456789

#1 h Mar 24, 2025, 11:17 AM

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Let $x, y$ be positive real numbers satisfying $x + y = 2$ . Prove that

$3(x^{\frac{2}{3}} + y^{\frac{2}{3}}) \geq 4 + 2x^{\frac{1}{3}}y^{\frac{1}{3}}.$

This post has been edited 1 time. Last edited by pennypc123456789, Mar 24, 2025, 11:22 AM

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#2 h Mar 24, 2025, 11:35 AM • 1 Y

Y by pennypc123456789

Sorry, fakesolved

This post has been edited 1 time. Last edited by User21837561, Mar 24, 2025, 11:46 AM
Reason: Wrong solve

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#4 h Mar 24, 2025, 12:57 PM • 1 Y

Y by pennypc123456789

Let $x = a^3, y = b^3$ , where $a^3 + b^3 = 2$ . Then, the problem becomes equivalent to $3(3a^2+3b^2-2ab)\ge 12$ where the constraint is given as $a^3+b^3=2$ . Using the fact that $2 = a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$ , we obtain these following identities
$2a^2+2b^2-2ab = 2 \cdot \frac{2}{ab} = \frac{4}{ab} \Longleftrightarrow 3a^2 + 3b^2 - 2ab = \frac{4}{a+b}+a^2+b^2$ $3a^2+3b^2-3ab=3 \cdot \frac{2}{ab} =\frac{6}{ab} \Longleftrightarrow 3a^2+3b^2-2ab=\frac{6}{a+b}+ab$ Thus, we have
$3(3a^2+3b^2-2ab)\ge12$ $\Longleftrightarrow 2(3a^2+3b^2-2ab)+(3a^2+3b^2-2ab)\ge12$ $\Longleftrightarrow 2 \left(\frac{6}{a+b}+ab\right)+\frac{4}{a+b}+a^2+b^2 \ge12$ $\Longleftrightarrow \frac{8}{a+b}+\frac{8}{a+b}+(a+b)^2\ge 12$ which is true by AM-GM.

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sqing

#5 h Apr 5, 2025, 12:15 PM

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Let $a,b$ be real numbers such that $a^3+b^3=2.$ Prove that
$3(a^2+b^2) \geq 4 + 2ab$ $a^2+b^2+a+b+\frac{8}{a+b}\geq 8$ $\frac{1}{a}+\frac{1}{b}\ge 2(a^2 - a + 1)(b^2 - b + 1)$ Let $a,b \geq 0$ and $a^3 +b^3=2.$ Prove that
$3(a^4+b^4)+2a^4b^4\leq 8$ h h h

This post has been edited 2 times. Last edited by sqing, Apr 5, 2025, 12:25 PM

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sqing

#6 h Apr 5, 2025, 12:39 PM

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Let $a,b \geq 0$ and $a^3 +b^3=2.$ Prove that
$a^2+b^2\leq 2$ Old
Let $a,b$ be real numbers such that $a^3 +b^3=27.$ Prove that
$a^2+b^2\geq 9$ Let $a,b$ be real numbers such that $3(a^4+b^4)+2a^4b^4\leq 8.$ Prove that
$a^2+b^2\leq 2$

This post has been edited 3 times. Last edited by sqing, Apr 5, 2025, 1:21 PM

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#7 h Apr 6, 2025, 3:12 AM

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sqing wrote:

Let $a,b$ be real numbers such that $a^3+b^3=2.$ Prove that
$3(a^2+b^2) \geq 4 + 2ab$ $a^2+b^2+a+b+\frac{8}{a+b}\geq 8$ $\frac{1}{a}+\frac{1}{b}\ge 2(a^2 - a + 1)(b^2 - b + 1)$ Let $a,b \geq 0$ and $a^3 +b^3=2.$ Prove that
$3(a^4+b^4)+2a^4b^4\leq 8$ h h h

Let $s=a+b$ and $t=ab$
$a^3 +b^3=2 \rightarrow s>0$ $a^3 +b^3=2 \iff t= \frac{s^3-2}{3s}$ $3(a^2+b^2) \geq 4 + 2ab \iff 3(s^2-2t) \geq 4+2t \iff s^3-12s+16 \geq 0$ $a^2+b^2+a+b+\frac{8}{a+b}\geq 8 \iff s^3+3s^2-24s+28 \geq 0$

This post has been edited 3 times. Last edited by SunnyEvan, Apr 6, 2025, 3:19 AM

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sqing

#11 h Apr 6, 2025, 1:03 PM

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Good.Thanks.

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