# Prove expression is positive

Let $\sum_i w_i=1$ and $w_i, x_i \in \mathbb{R}$, show that $$\sum_i w_i x_i^4-\sum_i w_i x_i \sum_i w_i x_i^3\geq 0.$$

I can show that $\sum_i w_i x^2 -\sum_i w_i x_i \sum_i w_i x_i \geq 0$ by convexity of the function $f(y)=y^2$, but i am not sure that this helps.

EDIT: Sorry i forgot: $w_i\geq0$.

• you need $w_i\ge 0$ for $\sum_i w_i x_i^2 \ge \left( \sum_i w_i x_i \right)^2$. – user251257 Mar 17 '16 at 18:02

Not true as stated. If $w_1 = -1$, $w_2 = 2$, $x_1 = 1$, $x_2 = 0$, the left side is $-2$.
If the $w_i$ are nonnegative, then recast the problem probabilistically: You have a random variable $X$ that takes value $x_i$ with probability $w_i$. Then the inequality is equivalent to $$E(X^4)-E(X)E(X^3)\ge0$$ which follows from the fact that $X$ and $X^3$ are positively (ok, not-negatively) correlated (draw a picture!), and therefore $$0\le \operatorname{Cov}(X,X^3) = E(XX^3)-E(X)E(X^3).$$ More rigorously, the general result is: If $h$ is a nondecreasing function, then $\operatorname{Cov}(X,h(X))\ge0$.
Proof: By definition, $$\operatorname{Cov}(X,h(X)):=E(X-EX)(h(X)-Eh(X)).$$ Write $$(X-EX)(h(X)-Eh(X))=(X-EX)(h(X)-h(EX)) + (X-EX)(h(EX)-Eh(X))\tag1$$ The first term on the RHS of (1) is a nonnegative random variable, since $h$ is nondecreasing. The second term has expectation zero since $h(EX)$ and $Eh(X)$ are both constant.
• how do you prove $\operatorname{Cov}(X,X^3)\ge 0$? – user251257 Mar 17 '16 at 19:04