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sqing

#1 h Mar 30, 2025, 1:09 PM

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Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $ad+bc\ge 6.$ Prove that
$0\leq abcd \leq 9$ $-\frac{13}{2}\leq ab+cd \leq \frac{13}{2}$ $-5\leq a+bc+d \leq \frac{169}{24}$ $-2 \leq a+b^2 \leq \frac{17}{4}$ $6 \leq a^2+bc+d^2 \leq 13$ $-2\sqrt 2 \leq a+b \leq 2\sqrt 2$

This post has been edited 1 time. Last edited by sqing, Mar 31, 2025, 4:34 AM

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sqing

#2 h Mar 31, 2025, 4:39 AM

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sqing wrote:

Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $ad+bc\ge 6.$ Prove that $-5\leq a^2+b+c+d^2 \leq \frac{701}{52}$

Solution of DAVROS:
Let $a=2\cos x, b=2\sin x, c=3\cos y, d=3\sin y$ and $f= a^2+b+c+d^2$

From condition $6\sin(x+y) \ge6 \implies x+y=\frac{\pi}2 \implies c=3\sin x, d=3\cos x, f= 13+5\sin x - 13\sin^2 x$

For max : $f= \frac{701}{52}-13\left( \sin x-\frac5{26} \right)^2 \le \frac{701}{52}$ @ $\sin x = \frac5{26}, \cos x = \pm\frac{\sqrt{651}}{26}$

For min : $f$ is concave in $\sin x$ so $f\ge \max \{ f_{\sin x =1}, f_{\sin x=-1}\}=-5$ @ $\sin x=-1, \cos x=0$

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