Prove $(a^p + b^p)^{1/p} \leq (a^2 + b^2)^{1/2}$ Given $a,b \geq 0$ and $2 \leq p <  \infty$ prove that $(a^p + b^p)^{1/p} \leq (a^2 + b^2)^{1/2}$. I tried using the convex function $f(x) = x^p$ and then prove $a^p + b^p \leq (a^2 + b^2)^{p/2}$ but can't seem to make any real progress. Can someone help me?
 A: Let $c=a/b$ and observe $f:[2,\infty)\to \mathbb{R}$ defined by $$f(x) := (c^2+1)^x-(c^x+1)^2 $$then $$f'(x)=(c^2+1)^x\ln(c^2+1)-(c^x+1)c^x\ln c^2$$
If $f'(x)>0$ then $f$ is increasing so $f(x)\geq 0$ for all $x$ since $f(2)=0$
Clearly $\ln(c^2+1)>\ln(c^2)$ so it is enough to check if $$(c^2+1)^x \geq c^x(c^x+1)$$ for all $x\geq 2$, i.e. $$(c+{1\over c})^x \geq c^x+1$$
Let $$g(x) := (c+{1\over c})^x - c^x- 1$$ Then $$g'(x) = (c+{1\over c})^x\ln (c+{1\over c}) - c^x\ln c$$
Since clearly $g'(x)\geq 0$ we are done.
A: A more general inequality is true ( probably already on this site, but could not find the duplicate)
If $a_i\ge 0$ and $0< x < y$ then
$$\left( \sum a_i^y\right)^{\frac{1}{y}} \le \left( \sum a_i^x\right)^{\frac{1}{x}}$$
Indeed, consider $a_i^x= b_i$, and $z = \frac{y}{x}$. We have to show that
$$\left(\sum b_i^z\right)^{\frac{1}{z}}\le \sum b_i$$
or
$$\sum \left(\frac{b_i}{\sum b_k}\right)^z \le 1$$
But now $c_i = \frac{b_i}{\sum b_k}$ are positive with sum $1$, and $z>1$, so
$$\sum c_i^z \le \sum c_i = 1$$
