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sqing

#1 h Apr 11, 2025, 2:57 AM • 1 Y

Y by cubres

Let $a,b,c,d\geq 0 ,a-b+d=21$ and $a+3b+4c=101$ . Prove that
$61\leq a+b+2c+d\leq \frac{265}{3}$ $- \frac{2121}{2}\leq ab+bc-2cd+da\leq \frac{14045}{12}$ $\frac{519506-7471\sqrt{7471}}{27}\leq ab+bc-2cd+3da\leq 33620$

This post has been edited 2 times. Last edited by sqing, Apr 11, 2025, 3:13 AM

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Bet667

#2 h Apr 11, 2025, 2:59 PM • 2 Y

Y by cubres, andyzhuang2010

By finding sum of two equationn dividing both sides by two gives us a+b+2c+d/2=64 and that proves the 61<= where inequality holds for d=0

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sqing

#3 h Apr 11, 2025, 3:49 PM • 2 Y

Y by Bet667, cubres

Good.Thanks.
$a+b+2c+\frac{d}{2}=61$ $a+b+2c+d=61+\frac{d}{2}\geq 61$

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kokcio

#4 h Apr 12, 2025, 5:31 PM • 1 Y

Y by cubres

$a+b+2c+d=61+\frac{d}{2}$ , so we want to prove $d\leq\frac{164}{3}$ , but $b=a+d-21$ , so $4a+4c+3d=164$ . Therefore, $d\leq\frac{164}{3}$ , because $a,c\geq0$ .

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kokcio

#5 h Apr 12, 2025, 5:51 PM • 1 Y

Y by cubres

Solving this equations for $a,b$ gives us $a=41-c-\frac{3d}{4}$ and $b=20-c+\frac{d}{4}$ . Therefore, $ab+bc-2cd+da=-\frac{9cd}{4}-41c-\frac{15d^2}{16}+\frac{145d}{4}+820$ .
But $-\frac{9cd}{4}-41c-\frac{15d^2}{16}+\frac{145d}{4}+820\leq-\frac{15d^2}{16}+\frac{145d}{4}+820\leq\frac{4205}{12}+820=\frac{14045}{12}$ , so $ab+bc-2cd+da\leq\frac{14045}{12}$ and equality holds for $(a,b,c,d)=(\frac{53}{2},\frac{149}{6},0,\frac{58}{3})$

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kokcio

#6 h Apr 12, 2025, 6:43 PM • 1 Y

Y by cubres

Let $f(x,y)=\frac{15y^2}{16}+\frac{9}{4}xy+41x-\frac{145}{4}y$ . This function is linear in $x$ and $\frac{9d}{4}+41>0$ .
We know that $a,b\geq0$ , so we have inequalities $c+\frac{3d}{4}\leq41$ and $c\leq20+\frac{d}{4}$ . Therefore, $f(c,d)\leq f(min\{41-\frac{3d}{4},20+\frac{d}{4}\},d)$ .
We can see that if $f(20+\frac{d}{4},d)>\frac{3761}{2}$ then $d>21$ and if $f(41-\frac{3d}{4},d)>\frac{3761}{2}$ then $d<21$ (just solve quadratic equation). This means that for all $d>0$ we have to have $f(min\{41-\frac{3d}{4},20+\frac{d}{4}\},d)\leq\frac{3761}{2}$ , so $f(c,d)\leq\frac{3761}{2}$ , but this is equivalent with that $ab+bc-2cd+da\geq-\frac{2121}{2}$ . Equality holds for $(a,b,c,d)=(0,0,\frac{101}{4},21)$

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sqing

#7 h Apr 13, 2025, 2:08 AM • 1 Y

Y by cubres

Nice.Thanks.

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