# Inequality with $abc=2$

Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $\frac{a+b+c}{4}\geq \sqrt{\frac{a^2+b^2+c^2}{6}}$ 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 $$a+b+c\geq\sqrt{81abc(a^2+b^2+c^2)}$$ so it's enough to prove that $$\left(\frac{\sqrt{81abc(a^2+b^2+c^2)}}{4}\right)^{12}\geq\left(\frac{a^2+b^2+c^2}{6}\right)^3\cdot\frac{a^2b^2c^2}{4},$$ which is wrong for $abc\rightarrow0^+$.

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. Aug 5, 2018 at 9:36
• You could expand the terms out and keep using the AM-GM inequality you could get answer. This however would take ages. The answer below is probably as good as you're going to get if you don't want to crunch it out using AM-GM (it's not very straightforward to apply AM-GM here, you have to group terms together but by bit to eventually make it work. ) Aug 5, 2018 at 12:58
• @Cataline I think AM-GM for three numbers does not help here because the equality does not occur for $a=b=c.$ Aug 6, 2018 at 4:29

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 $a=b=\sqrtt$.

Thus, $c=\frac{2}{\sqrt{t^2}}$ and we need to prove that $$\frac{2\sqrtt+\frac{2}{\sqrt{t^2}}}{4}\geq\sqrt{\frac{2\sqrt{t^2}+\frac{4}{\sqrt{t^4}}}{6}}$$ or $$\frac{t+1}{2}\geq\sqrtt\sqrt{\frac{t^2+2}{3}}$$ 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!

• Thank you very much. I think this is not a nice solution. Aug 5, 2018 at 10:00
• Try to find something better! Aug 5, 2018 at 10:01
• I am sure that it's impossible to find something better than my solution. Good luck! Aug 5, 2018 at 10:07
• Hi, @MichaelRozenberg. I think this is a good first step but there's some parts I'm not sure about. How do you justify that $f(u)$ is increasing? If we fix $v^2=1.8$, we can show that $(u,w,v)$ is admissible (i.e. $T(u,v,w)\geq0$) iff $1.403\leq u\leq1.475$. But then we can show that over the interval $1.403\leq u\leq1.475$, $f(u)$ is decreasing, not increasing.
– Jam
Aug 6, 2018 at 11:44
• @Jam I said about $f(v^2)$ and not about $f(u)$. It was typo. I fixed. Thank you! Aug 6, 2018 at 11:47

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.

• $(a+b+c)^2\geq a^2+b^2+c^2$??? Aug 5, 2018 at 10:27