Generalized Hölder inequality: Application

Question

how can I prove the following inequality (we can use Hölder inequality) ?

With quantities all positive and $n\geq 1$ we have

$$ (a+b)^{n-1}(A+B) \geq ((a^{n-1}A)^{1/n} + (b^{n-1}B)^{1/n})^n $$

thank you very much

Answer 1

It's just Holder.

The Holder inequality it's the following:

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $\alpha$ and $\beta$ be positive numbers. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

In our case $a_1=a$, $a_2=b$, $b_1=A$, $b_2=B$, $\alpha=n-1$ and $\beta=1.$

A proof of Holder.

Let $f(x)=x^k$, where $k>1$.

Thus, $f$ is a convex function and by Jensen we obtain: $$\alpha_1x_1^k+\alpha_2x_2^k+...+\alpha_nx_n^k\geq(\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n)^k,$$ where $\alpha_i>0$ such that $\alpha_1+\alpha_2+...+\alpha_n=1$ and $x_i>0$, or $$(\alpha_1+\alpha_2+...+\alpha_n)^{k-1}\left(\alpha_1x_1^k+\alpha_2x_2^k+...+\alpha_nx_n^k\right)\geq(\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n)^k.$$ Now, let $k=1+\frac{\alpha}{\beta}$ for $\alpha>0$ and $\beta>0$, $\alpha_i=a_i$ and $\alpha_ix_i^k=b_i.$

We obtain: $$(a_1+a_2+...+a_n)^{\frac{\alpha}{\beta}}(b_1+b_2+...+b_n)\geq$$ $$\geq\left(a_1\left(\frac{b_1}{a_1}\right)^{\frac{1}{1+\frac{\alpha}{\beta}}}+a_2\left(\frac{b_2}{a_2}\right)^{\frac{1}{1+\frac{\alpha}{\beta}}}+...+a_n\left(\frac{b_n}{a_n}\right)^{\frac{1}{1+\frac{\alpha}{\beta}}}\right)^{1+\frac{\alpha}{\beta}}$$ or $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$ and we are done!