Inequality with $abc=2$

Question

Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $\frac{a+b+c}{4}\geq \sqrt[4]{\frac{a^2+b^2+c^2}{6}}$ Sorry, I don't have an idea for this problem.If I had, I would have showed them all. We need to prove that: $$\left(\frac{a+b+c}{4}\right)^{12}\geq\left(\frac{a^2+b^2+c^2}{6}\right)^3\cdot\frac{a^2b^2c^2}{4}.$$ Now, by the Vasc's inequality $(a+b+c)^5\geq81abc(a^2+b^2+c^2)$ We can obtain $$a+b+c\geq\sqrt[5]{81abc(a^2+b^2+c^2)}$$ so it's enough to prove that $$\left(\frac{\sqrt[5]{81abc(a^2+b^2+c^2)}}{4}\right)^{12}\geq\left(\frac{a^2+b^2+c^2}{6}\right)^3\cdot\frac{a^2b^2c^2}{4},$$ which is wrong for $abc\rightarrow0^+$.

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, our inequality is equivalent to $f(v^2)\geq0$, where $$f(v^2)=\frac{3}{4}u-\left(\frac{3u^2-2v^2}{2}\right)^{\frac{1}{4}}\left(\frac{w^3}{2}\right)^{\frac{1}{6}}.$$ We see that $f$ increases, which says that it remains to prove our inequality for a minimal value of $v^2$, which happens for an equality case of two variables.

Now, let $a=b=\sqrt[3]t$.

Thus, $c=\frac{2}{\sqrt[3]{t^2}}$ and we need to prove that $$\frac{2\sqrt[3]t+\frac{2}{\sqrt[3]{t^2}}}{4}\geq\sqrt[4]{\frac{2\sqrt[3]{t^2}+\frac{4}{\sqrt[3]{t^4}}}{6}}$$ or $$\frac{t+1}{2}\geq\sqrt[3]t\sqrt[4]{\frac{t^2+2}{3}}$$ or $g(t)\geq0,$ where $$g(t)=\ln\frac{t+1}{2}-\frac{1}{3}\ln{t}-\frac{1}{4}\ln\frac{t^2+2}{3}.$$ But $$g'(t)=\frac{1}{t+1}-\frac{1}{3t}-\frac{t}{2(t^2+2)}=\frac{(t-2)^2(t-1)}{6t(t+1)(t^2+2)},$$ which gives $t_{min}=1$ and since $g(1)=0,$ we are done!

Answer 2

I think @Michael_Rozenberg's answer is one of the best solutions to this problem you'll find but I found it quite confusing to follow since I didn't know about uvw substitution. So I'd like to attempt to clarify some points.

At first, it seems unjustified to be able to set a function of all three variables $u,w,v$ to a function of only one variable, $f(v^2)$, but you can accomplish this by fixing $u$ and $w$ to specific values. It is then possible to vary $v$, while keeping the values of $u$ and $w$ constant. In fact $w$ is already a constant due to the rule that $abc=2$.

We can show that $f(v^2)$ is strictly increasing, when $u$ and $w$ are fixed by computing the derivative of $f(v^2)$. Indeed, we can use the chain rule to show that…

$$f(x)=\frac{3}{4}u-\left(\frac{3u^2-2x}{2}\right)^{1/4}\left(\frac{w^3}{2}\right)^{1/6}\\ =\frac{3}{4}u-\left(\frac{3u^2}{2}-x\right)^{1/4}\\ \Rightarrow f'(x) =-\frac14\left(\frac{3u^2}{2}-x\right)^{-3/4}\left(-1\right)\\ =\frac{1}{4\left(\frac{3u^2}{2}-x\right)^{3/4}}$$

I've replaced $v^2$ by $x$ because it makes the algebra clearer. The term $\left(\frac{3u^2}{2}-x\right)^{3/4}$ can never be negative for real $x$, hence $f'(x)\geq0$ and $f(x)$ is strictly increasing.

We then want to find the value of $v^2$ such that $f(v^2)$ is minimal, which will show us that for all $v^2\in\mathbb{R^+}$, $f(v^2)\geq f(v_{\min}^2)$ and whether $f(v^2)\geq0$. We can use a corollary of Tejs' theorem to help us here:

Any symmetric polynomial of degree $\leq5$ in nonnegative real variables $a,b,c$ with a global minimum and/or maximum will attain this value at triples $(a,b,c)$ with either two of the variables equal or one of the variables equal to zero. Source: brilliant.org.

$ab+ac+bc$ is a symmetric polynomial of degree $2$, so we are permitted to use the theorem when minimising $f(v^2)$.

The rest of the proof follows from elementary algebra. If you'd like more help with the uvw method, please see this page.