If a,b>0 then the minimum value of y=b2a−x+a2x,0<x<a is

y=b2a−x+a2x≥2b2a2(a−x)x=2ab(a−x)x(useAM≥GM)Dr is max ie ax–x2 is max formula4ac−b24a=4(−1)0−a24(−1)=a24∴y=2aba2=4b

If a,b>0 then the minimum value of y=b2a−x+a2x,0<x<a is

y=b2a−x+a2x≥2b2a2(a−x)x=2ab(a−x)x(useAM≥GM)Dr is max ie ax–x2 is max formula4ac−b24a=4(−1)0−a24(−1)=a24∴y=2aba2=4b

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If $a, b > 0$ then the minimum value of $y = \frac{b^{2}}{a - x} + \frac{a^{2}}{x}, 0 < x < a$ is

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$y = \frac{b^{2}}{a - x} + \frac{a^{2}}{x} \geq 2 \sqrt{\frac{b^{2} a^{2}}{(a - x) x}}$

$= \frac{2 a b}{\sqrt{(a - x) x}} (use AM \geq GM)$

Dr is max ie $a x - x^{2}$ is max formula

$\frac{4 a c - b^{2}}{4 a} = \frac{4 (- 1) 0 - a^{2}}{4 (- 1)} = \frac{a^{2}}{4}$

$∴ y = \frac{2 a b}{\frac{a}{2}} = 4 b$

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If $a, b > 0$ then the minimum value of $y = \frac{b^{2}}{a - x} + \frac{a^{2}}{x}, 0 < x < a$ is