# Prove $\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt{\frac{xyz}{16}}$

Let $x,y,z>0$. Prove that $$\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt{\frac{xyz}{16}}$$

On the left-hand side we have arithmetic and harmonic means, while on the right geometric mean. The AM-GM-HM inequality states that AM $\geq$ GM $\geq$ HM, but here HM is on the wrong side of the inequality.

## 2 Answers

Since we have a homogeneous inequality, we may assume $xyz=1$ without loss of generality, then find the infimum of $AM(x,y,z)+HM(x,y,z)$ over the set $$D=\left\{(x,y,z)\in\mathbb{R}_{+}^3: xyz=1\right\}.$$ So we want: $$\inf_{\mathbb{R}^+\times\mathbb{R}^+}\left(\frac{x+y+\frac{1}{xy}}{3}+\frac{3}{xy+\frac{1}{x}+\frac{1}{y}}\right)=\inf_{\mathbb{R}^+\times\mathbb{R}^+}\left(\frac{x^2 y+xy^2+1}{3xy}+\frac{3xy}{x^2y^2+x+y}\right).$$ Let $f(x,y)$ be the last function. It is a symmetric function, hence it is enough to study its behaviour over the set $0<y\leq x$. So, let $y=kx$ with $k\in(0,1]$. We have: $$g_k(x)=f(x,kx) = \frac{1}{3kx^2}+\frac{(k+1)x}{3}+\frac{3kx}{1+k+k^2 x^3}$$ and disregarding the last (positive) term, by the AM-GM inequality: $$g_k(x)\geq \frac{1}{3kx^2}+\frac{(k+1)x}{6}+\frac{(k+1)x}{6} \geq 3\sqrt{\frac{(k+1)^2}{108k}}$$ where the last term is a decreasing function over $(0,1]$. We just need a little improvement of the last inequality to prove the original claim. So, let us study $f(x,y)$ under another perspective. Assuming $x+y=k$ we have: $$f(x,y) = \frac{k+\frac{1}{xy}}{3}+\frac{3}{\frac{k}{xy}+xy}=\frac{k+\frac{1}{x(k-x)}}{3}+\frac{3}{\frac{k}{x(k-x)}+x(k-x)}$$ and by putting $x=\lambda k, y=(1-\lambda)k$ we are left with: $$\frac{k+\frac{1}{k^2\lambda(1-\lambda)}}{3}+\frac{3}{\frac{1}{k\lambda(1-\lambda)}+k^2\lambda(1-\lambda)}=\color{red}{\frac{k+\frac{1}{k^2 L}}{3}+\frac{3}{\frac{1}{kL}+k^2 L}=h(L,k)}$$ where $L$ ranges over $\left(0,\frac{1}{4}\right)$ and $k$ ranges over $\mathbb{R}_+$. The stationary points of $h(L,k)$ for a fixed $k$ occur at $$L \in\left\{\pm\frac{1}{\sqrt{2} k^{3/2}},\pm\frac{1}{\sqrt{5} k^{3/2}}\right\}$$ and $$h\left(\frac{1}{\sqrt{2} k^{3/2}},k\right) = \frac{4}{3}\sqrt{\frac{2}{k}}+\frac{k}{3},\qquad h\left(\frac{1}{\sqrt{5} k^{3/2}},k\right) = \frac{5}{6}\sqrt{\frac{5}{k}}+\frac{k}{3},$$ so the AM-GM inequality proves the claim.

In facts, the original inequality is optimal and our approach shows that equality is attained by $$\color{red}{(x,y,z) = \lambda\cdot (1,4,4)}$$ and cyclic shifts.

• @Lucian: where? After defining $D$, I have eliminated $z$ through the constraint $xyz=1$. – Jack D'Aurizio May 30 '16 at 21:27
• Have you also tried tackling this question ? – Lucian May 30 '16 at 21:38
• @Lucian: I will think about that, but non-homogeneous inequalities usually requires different tricks than a suitable change of variables plus partial derivatives. – Jack D'Aurizio May 30 '16 at 22:21
• Hello Jack. Is there any book you reccomend where to learn about inequalities (homogenous or non-homogenous)? I heard "homogenous inequality" before but I never knew why "homogenous" is important; now thanks to your answer I see some of the tricks you can use, and I would like to learn more – Ovi May 25 '18 at 4:36
• @Ovi: I recommend the books from Micheal Steele (The Cauchy-Schwarz master class) and Pham Kim Hung (I do not recall the exact title at the moment, but it should not be too difficult to find). The former is a bit more argumentative, the latter a bit more concise and technical, but they are both good sources of knowledge about the topic. – Jack D'Aurizio May 25 '18 at 12:56

Let $$x+y+z=3u$$, $$xy+xz+yz=3v^2$$ and $$xyz=w^3$$.

Thus, we need to prove that $$f(u)\geq0,$$ where $$f(u)=u+\frac{w^3}{v^2}-\frac{5w}{\sqrt{16}}.$$ Id est, it's enough to prove our inequality for the minimal value of $$u$$, which happens for equality case of two variables.

Since our inequality is symmetric and homogeneous, we can assume $$y=z=1$$ and $$z=16a^3$$

and we need to prove that: $$\frac{16a^3+2}{3}+\frac{48a^3}{32a^3+1}\geq5a$$ or $$(4a-1)^2(32a^4+16a^3-24a^2+a+2)\geq0$$ or $$(4a-1)^2(2(4a^2-1)^2+a(4a-1)^2)\geq0,$$ which is obvious.