Math, asked by Anonymous, 9 months ago

If a,b are positive real numbers such that a−b=2, then find the smallest value of the constant L for which $\sqrt{x^2+ax}$ - $\sqrt{x^2+bx}$
<1 for all x>0.

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Answered by Anonymous

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If a,b are positive real numbers such that a−b=2, then find the smallest value of the constant L for which -<1 for all x>0.​

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If a,b are positive real numbers such that a−b=2, then find the smallest value of the constant L for which $\sqrt{x^2+ax}$ - $\sqrt{x^2+bx}$
<1 for all x>0.