# Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$?

If $$x,y,z$$ are positive real numbers such that $$x+y+z=1$$ then is it true that

$$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$$ ?

• @miracle173 No, it isn't. The sum is $1$, and the product is $4*4*4=64$. – Henrik Nov 23 '14 at 8:55
• It smells like Jensen's inequality. – carlosayam Nov 23 '14 at 8:57
• Hum, Jensen's inequality seems to give an smaller bound ... sorry. – carlosayam Nov 23 '14 at 9:08

By inequality between harmonic and geometric mean we have: $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\ge\left(\frac{3}{\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}}\right)^3$$ Now if we prove that $$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\le\frac{3}{4}$$ we are done. But it is the same to prove that $$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{3}{\frac{x+1+y+1+z+1}{3}}=\frac{9}{4}.$$ The last inequality is the inequality between harmonic and arithmetic mean.

By using Jensen's inequality with the convex function $f(x) = \log(1+\frac1x)$, we get $$f(x)+f(y)+f(y) \ge 3f\left(\frac{x+y+z}3\right) = 3\log4$$ which on exponentiation gives the inequality.

• Very economical.+1 – daniel Nov 23 '14 at 10:20

By AM-GM, we have $$1+x=\frac{1}3+\frac13+\frac13+x\ge 4\sqrt[4]{\frac{x}{3^3}}.$$ So $$(1+x)(1+y)(1+z)\ge 4^3\sqrt[4]{\frac{xyz}{3^9}}.$$ We then need only show that $$\sqrt[4]{\frac{xyz}{3^9}}\ge xyz,$$ or that $$1\ge 3^3xyz.$$ The last inequality is clear thanks to AM-GM.

It is equivalent to show $$(x+x+y+z)(x+y+y+z)(x+y+z+z)\geq 64 xyz.$$

This is just AM-GM.

Use AM-GM to $(x+x+y+z)$, $(x+y+y+z)$, $(x+y+z+z)$.

Use the first equation to rewrite $z$in terms of $x$ and $y$, and then consider Consider the function $\mathbb R^2\to\mathbb R$: $$f(x,y)= \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{\text{your expression for z}}\right)$$ you are then looking for it's minimum which should be findable by elementary analytic methods.