Minimum Value? Hint (AM-GM)

Algebra Level 2

Find the minimum value of e cos ⁡ ( 2 x ) + e 2 − 2 cos ⁡ ( 2 x ) − 2 sin ⁡ 2 ( x ) \large e^{\cos(2x)} + e^{2 - 2\cos(2x) - 2\sin^2(x)} for real x x .

e \sqrt{e} 2 e 2 2e^{2} 2 e 2 \sqrt{e} e 2 e^{2}

Barry Leung by Barry Leung , 17, Hong Kong

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1 solution

Toby M
Jul 13, 2020

By AM-GM, we know that a + b ≥ 2 a b a + b ≥ 2 \sqrt{ab} . Now if a = exp ⁡ ( cos ⁡ 2 x ) , b = exp ⁡ ( 2 − 2 cos ⁡ 2 x − 2 sin ⁡ 2 x ) a = \exp \left(\cos 2x \right), b = \exp \left(2 - 2 \cos{2x} - 2 \sin^2 x \right) , then:

a b = exp ⁡ ( cos ⁡ 2 x + 2 − 2 cos ⁡ 2 x − 2 sin ⁡ 2 x ) = exp ⁡ ( − cos ⁡ 2 x + 2 − 2 sin ⁡ 2 x ) = exp ⁡ ( − cos ⁡ 2 x + sin ⁡ 2 x + 2 − 2 sin ⁡ 2 x ) = exp ⁡ ( − cos ⁡ 2 x + − sin ⁡ 2 x + 2 ) = exp ⁡ ( − 1 + 2 ) = e \displaystyle ab = \exp \left(\cos 2x + 2 - 2 \cos{2x} - 2 \sin^2 x \right) = \exp \left(-\cos 2x + 2 - 2 \sin^2 x \right) = \exp \left(-\cos^2 x + \sin^2 x + 2 - 2 \sin^2 x \right) = \exp \left(-\cos^2 x + -\sin^2 x + 2 \right) = \exp(-1+2) = e .

Therefore the minimum value of the expression must be 2 a b = 2 e 2 \sqrt{ab} = \boxed{2 \sqrt e} .

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Another Hong Kong student!

Barry Leung - 11 months ago

Yeah, it's not often you meet another Hong Konger on this site.

Toby M - 11 months ago

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