Focus on AM-GM

Algebra Level 4

Given that x ≥ 1 ; y ≥ 2 ; z ≥ 3 x\geq1; y\geq2 ; z\geq3 , let h be the maximum value of I I where I = y z x − 1 + x z y − 2 + x y z − 3 x y z . I=\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}. Submit your answer as ⌈ h ⌉ + 10 \lceil{h}\rceil+10 .

Part of the set


The answer is 12.

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P C by P C , 20, Vietnam

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1 solution

P C
Jan 27, 2016

Simplified I and we get I = x − 1 x + y − 2 y + z − 3 z I=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z} Using AM-GM, we get x − 1 x ≤ 1 \frac{\sqrt{x-1}}{x}\leq1 y − 2 y ≤ 1 2 2 \frac{\sqrt{y-2}}{y}\leq\frac{1}{2\sqrt{2}} z − 3 z ≤ 1 2 3 \frac{\sqrt{z-3}}{z}\leq\frac{1}{2\sqrt{3}} So I ≤ 1 + 1 2 2 + 1 2 3 = h ≈ 1.64 I\leq1+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=h\approx1.64 ⇒ ⌈ h ⌉ + 10 = 12 \Rightarrow\lceil{h}\rceil+10=12 The equality holds when ( x ; y ; z ) = ( 2 ; 4 ; 6 ) (x;y;z)=(2;4;6)

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@Gurīdo Cuong sir in which numbers have you applied am-gm

Deepansh Jindal - 4 years, 10 months ago

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please help sir

Deepansh Jindal - 4 years, 10 months ago

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