Let $A_0 , B_0 , A_1, B_1 \geqslant 0 $ and let $0 \leqslant a,b <1 $. Then prove that $ A_0^a B_0^b + A_1 ^a B_1 ^b \leqslant (A_0 + A_1 )^a (B_0 + B_1 )^b$
$ A_0^a B_0^b + A_1 ^a B_1 ^b \leqslant (A_0 + A_1 )^a (B_0 + B_1 )^b$ given $A_0 , B_0 , A_1, B_1 \geqslant 0 $ ; 0 \leqslant a,b <1
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calculus
inequality
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0except the case of $0^0$. – 2012-06-17
1 Answers
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$A_0 = A_1 = B_0 = B_1 = x$ and $a+b < 1$. Then $A_0^a B_0^b + A_1^a B_1^b = 2x^{a+b}$ $(A_0 + A_1)^a (B_0 + B_1)^b = (2x)^{a+b}$ Since $a+b < 1$, we have $(A_0 + A_1)^a (B_0 + B_1)^b = (2x)^{a+b} < 2x^{a+b} = A_0^a B_0^b + A_1^a B_1^b$ contradicting your claim.