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Mathematics, 21.06.2021 14:00 andrew763

i will mark brainliest, thank, and 5star Find the smallest positive value of the nonnegative number \lambdaλ such that the inequality
\frac{a+b}{2}\ge \lambda \sqrt{ab}+(1-\lambda) \sqrt{\frac{a^2+b^2}{2}}
2
a+b

≥λ
ab

+(1−λ)
2
a
2
+b
2





holds for all positive real numbers a, ba, b.

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