Generalization of MYM , April 2018 Q4

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Generalization of MYM , April 2018 Q4

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Source: Zhangyanzong

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sqing

#1 h May 11, 2018, 3:34 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b$ be real numbers such that $a^3+b^3=2$ . Prove that
$a^2+b^2+\frac{k}{a+b}\geq\left(\frac{3k+4}{2}\right)^{\frac{2}{3}},(-\frac{4}{3}<k\leq 4)$ $a^2+b^2+\frac{k}{a+b}\geq\frac{k+4}{2},(k> 4)$ Mathematical and Youth Magazine Issue 490 , April 2018 ,Problem 4

This post has been edited 1 time. Last edited by sqing, May 11, 2018, 3:39 AM

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#2 h May 11, 2018, 9:21 AM • 1 Y

Y by Adventure10

hi
for the second one : $a^3+b^3=(a+b)^3-3ab(a+b)\ge (a+b)^3-3/4(a+b)^3=\frac{1}{4}$
Thus , $8 \ge (a+b)^3$
So $2 \ge a+b$
By easy computations the inequality given id equivalent to $(2-a-b)(k+4-2ab)\ge 4-4ab$
$2-a-b$ is positive so $k+4-2ab\ge k+2$
i will come with a full solution hopefully soon

This post has been edited 3 times. Last edited by goldenturtle, May 11, 2018, 9:24 AM

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sqing

#3 h May 12, 2018, 1:01 AM • 1 Y

Y by Adventure10

sqing wrote:

Let $a,b$ be real numbers such that $a^3+b^3=2$ . Prove that
$a^2+b^2+\frac{k}{a+b}\geq\left(\frac{3k+4}{2}\right)^{\frac{2}{3}},(-\frac{4}{3}<k\leq 4)$ $a^2+b^2+\frac{k}{a+b}\geq\frac{k+4}{2},(k> 4)$ Mathematical and Youth Magazine Issue 490 , April 2018 ,Problem 4

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#4 h May 12, 2018, 8:29 PM • 2 Y

Y by Adventure10, Mango247

hi sqing
citing the previous chapter
$(3a^2-2ab+3b^2)^3 = \left( \frac{(a^2+2ab+b^2)+2(4a^2-4ab+4b^2)}{3} \right)^3 \ge (a^2+2ab+b^2)(4a^2-4ab+4b^2)^2 = 16(a^3+b^3)^2 = 64$
thus , $3a^2-2ab+3b^2\ge 4$
using the fact that $a+b=\frac{2}{a^2-ab+b^2}$ our inequality is equivalent to
$2a^2+2b^2+k(a^2-ab+b^2)\ge k+4$
so we have to prove that $(k+2)a^2+(k+2)b^2-kab\ge k+4$
which is equivalent to $6a^2+6b^2-4ab+(k-4)(a^2-ab+b^2)\ge k+4$
we already know that $6a^2+6b^2-4ab\ge 8$ so we only have to prove that $(k-4)(a^2-ab+b^2)\ge k-4$
since $k>4$ we can divide the inequality by $k-4$ resulting $a^2-ab+b^2=\frac{2}{a+b}\ge 1$
equivalent to $2\ge a+b$
$a^3+b^3=(a+b)^3-3ab(a+b)\ge (a+b)^3-3/4(a+b)^3=\frac{1}{4}(a+b)^3$
so $8\ge(a+b)^3$ so $2\ge(a+b)$
we are done .
P.S sorry for taking too long

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#5 h May 13, 2018, 1:15 AM • 2 Y

Y by Adventure10, Mango247

Thanks

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