How can i verify $(a+b)^\frac{1}{n} \le a^\frac{1}{n} + b^\frac{1}{n}$ for all 'a' and 'b'. are positive real no. I am trying to solve this problem. But I do not have any perfect proof of that.
 A: Why don't you raise both sides to the $n$-th power? Then you need to prove that 
$(a+b) \leq \left( a^{\frac{1}{n}} + b^{\frac{1}{n}} \right)^n = a + b~+$ positive terms, which is always true.
A: Considering $a,b,n$ as positive integers.
$ \begin{align*} a+b & \le a+\left[\dbinom{n}{1} a^{\frac{n-1}{n}} b^{\frac{1}{n}}+\cdots+\dbinom{n}{n-1}a^{\frac{1}{n}}b^{\frac{n-1}{n}}\right]+b \\ & \le \left( a^{\frac{1}{n}}+b^{\frac{1}{n}}\right)^n \\  (a+b)^{\frac{1}{n}}&\le a^{\frac{1}{n}}+b^{\frac{1}{n}} \end{align*} $
A: Write $x=a^{1/n}$ and $y=b^{1/n}$. Now prove $$x^n+y^n\leq (x+y)^n$$
which you can do by using the binomial theorem on right the side...
A: If $a,b>0$ and $n>1$ proving that $$(a+b)^{1/n} \le (a^{1/n}+b^{1/n})~~~(1)$$
Let $p=1/n<1,$ then
$$f(x)=(1+x)^p-1+x^p \Rightarrow f'(x)=p(1+x)^{p-1}-px ^{p-1}=px^{p-1}[(1+1/x)^{p-1}-1])<0, x \ge 0.$$ So $f(x)$ is a decreasing function for $x\ge 0$, then $$f(x) \le f(0)  \Rightarrow (1+x)^p-1-x^p \le 0 \Rightarrow  (1+(a/b))^p \le 1+(a/b)^p$$
and hence the result, the sign of the inequality reverses if $n<1$.
