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If $u={\sqrt {a^2\cos^2\alpha + b^2\sin^2\alpha}} + {\sqrt {a^2\sin^2\alpha + b^2\cos^2\alpha}}$, find the difference between the maximum and minimum value of $u^2$.

I tried squaring the expression on both sides but i am ending up with some complex expression, i.e, $a^2 + b^2 + 2\sin \alpha\cos \alpha\sqrt {(a^2-b^2)^2 + (a^2b^2)/(\sin^2 \alpha\cos^2 \alpha)}.$

Please Help.

José Carlos Santos
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Yami Kanashi
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1 Answers1

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By C-S $$u^2=a^2+b^2+2\sqrt{(a^2\cos^2\alpha+b^2\sin^2\alpha)(b^2\cos^2\alpha+a^2\sin^2\alpha)}\geq$$ $$\geq a^2+b^2+2\sqrt{(ab\cos^2\alpha+ab\sin^2\alpha)^2}=a^2+b^2+2|ab|=(|a|+|b|)^2.$$ The equality occurs when $$(a\cos\alpha,b\sin\alpha)||(b\cos\alpha,a\sin\alpha),$$ which says that $(|a|+|b|)^2$ is a minimal value.

Michael Rozenberg
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