For 0 leq p leq 1 and for any positive a, b, | vedclass

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(B) Given $0 \leq p \leq 1$ and $a, b > 0$. Consider the function $f(x) = x^p$. Since $0 \leq p \leq 1$, the function $f(x) = x^p$ is a concave functi

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$I(p) \leq J(p)$

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(B) Given $0 \leq p \leq 1$ and $a, b > 0$. Consider the function $f(x) = x^p$. Since $0 \leq p \leq 1$, the function $f(x) = x^p$ is a concave function for $x > 0$. By the property of concave functions, for any $a, b > 0$ and $0 This implies $I(p) \leq J(p)$. For $p=0$, $I(0) = (a+b)^0 = 1$ and $J(0) = a^0 + b^0 = 1 + 1 = 2$, so $1 \leq 2$. For $p=1$, $I(1) = a+b$ and $J(1) = a+b$, so $a+b \leq a+b$. Thus, $I(p) \leq J(p)$ holds for all $0 \leq p \leq 1$.

For $0 \leq p \leq 1$ and for any positive $a, b$, let $I(p)=(a+b)^{p}$ and $J(p)=a^{p}+b^{p}$. Then:

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