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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: $w_i\geq0$.

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    $\begingroup$ you need $w_i\ge 0$ for $\sum_i w_i x_i^2 \ge \left( \sum_i w_i x_i \right)^2$. $\endgroup$ – user251257 Mar 17 '16 at 18:02
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Not true as stated. If $w_1 = -1$, $w_2 = 2$, $x_1 = 1$, $x_2 = 0$, the left side is $-2$.

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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.

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  • $\begingroup$ how do you prove $\operatorname{Cov}(X,X^3)\ge 0$? $\endgroup$ – user251257 Mar 17 '16 at 19:04
  • $\begingroup$ @user251257 I've edited my answer to prove the general result. $\endgroup$ – grand_chat Mar 17 '16 at 19:30
  • $\begingroup$ ah I see. Nice trick. $\endgroup$ – user251257 Mar 17 '16 at 19:34

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