# AM/GM inequalities

I need some help to prove this inequality... I guess one can use Jensen's then AM/GM inequalities.

Let $$x_1, x_2, x_3, x_4$$ be non- negative real numbers such that $$x_1 x_2 x_3 x_4 =1$$.
We want to show that $$x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge x_1+x_2+x_3+x_4,$$ and also $$x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge \frac1{x_1}+ \frac1{x_2} +\frac1{x_3}+\frac1{x_4}.$$

Since $$x↦x^3$$ is convex on R+ by Jensen's Inequality we have $$x_1^3+x_2^3+x_3^3+x_4^3≥4^{-2}(x_1+x_2+x_3+x_4)^3$$ then using AM/GM inequality and since $$x_1x_2x_3x_4=1,$$ we can show that $$(x_1+x_2+x_3+x_4)^3≥4$$

• The second: multiply RHS by $x_1x_2x_3x_4$ and then apply Muirhead. The first should be killable in this way too: square both parts, multiply RHS by $x_1x_2x_3x_4$ and then apply Muirhead. Jul 17, 2020 at 14:14
• @Clifford I solved your problems. If you want to see my solutions show please your attempts. Jul 17, 2020 at 18:19
• @MichaelRozenberg Since $x \mapsto x^3$ is convex on $\mathbb{R}^+$ by Jensen's Inequality we have $x_1^3 + x_2^3 + x_3^3 + x_4^3 \ge 4^{-2} (x_1 + x_2 + x_3 + x_4)^3$ then using AM/GM inequality and since $x_1 x_2 x_3 x_4 = 1$, we can show that $(x_1 + x_2 + x_3 + x_4)^3 \ge 4$ Jul 17, 2020 at 18:43

$$\displaystyle\prod_{i=1}^n x_i=1,x_i\in\mathbb R^+$$. Show that

• $$(a)\displaystyle\sum_{i=0}^n x_i^3\ge \displaystyle\sum_{i=0}^nx_i$$
• $$(b)\displaystyle\sum_{i=0}^n x_i^3\ge \displaystyle\sum_{i=0}^n\frac{1}{x_i}$$

$$(a)$$ Let $$f(x)=x^3-x$$. Since $$\frac{d^2}{dx^2}f(x)=6x>0\ \forall\ x>0$$, Jensen's inequality $$\Rightarrow \dfrac{\displaystyle\sum _{i=1}^n(x^3_i-x_i)}n\ge f\left(\frac{\displaystyle\sum x_i}n\right)\ge f\left(\left(\displaystyle\prod_{i=1}^n x_i\right)^{\frac1n}\right)=f(1)=0\tag{1}$$ since $$f(x)$$ is increasing after $$x=1$$.

PS: Draw graph of $$f(x)$$ to understand it better.

• $\frac{d}{dx}x^{-3}<0$ for $x>0$. Also I think you're missing a cube around the right side of the inequality in $(a)$ when using Jensen. That conclusion is clearly false without some kind of further assumption. E.g. take $x_1=\ldots=x_n=\frac{1}{2}$. Jul 17, 2020 at 15:18
• It's highly doubtful that $(x^{-3})'>0$ as $(x^{-3})'=\color{red}-3x^{-4}$ Jul 17, 2020 at 15:56
• @halrankard I don't know what has happened to my eyes today. It's double derivative. Sorry! Jul 17, 2020 at 15:57
• @AlexeyBurdin Is my solution correct now? I am afraid because it's been a while since I last did questions like these... Jul 17, 2020 at 16:05
• I assume that's the inflection point where $x^3-x^{-1}$ becomes convex? Then, yes I agree. Jul 17, 2020 at 19:13

The first inequality.

We need to prove that: $$\sum_{cyc}x_1^3\geq\sum_{cyc}x_1\sqrt{\prod_{cyc}x_1}$$ or $$\sum_{cyc}x_1^3\geq\sum_{cyc}\sqrt{x_1^3x_2x_3x_4},$$which is true by Muirhead because $$(3,0,0,0)\succ(1.5,0.5,0.5,0.5).$$ We can use also AM-GM: $$\sum_{cyc}x_1^3=\frac{1}{6}\sum_{cyc}(3x_1^3+x_2^3+x_3^3+x_4^3)\geq\frac{1}{6}\sum_{cyc}6\sqrt{x_1^9x_2^3x_3^3x_4^3}=\sum_{cyc}x_1.$$ Also, the Tangent Line method works: $$\sum_{cyc}(x_1^3-x_1)=\sum_{cyc}\left(x_1^3-x_1-2\ln{x_1}\right)\geq0.$$ Indeed, let $$f(x)=x^3-x-2\ln{x},$$ where $$x>0$$.

Thus, $$f'(x)=3x^2-1-\frac{2}{x}=\frac{(x-1)(3x^2+3x+2)}{x},$$ which gives $$x_{min}=1$$, $$f(x)\geq f(1)=0$$ and we are done!

The second inequality we can prove by similar ways.