On positive reals if $3x+4y+7z=1$ what is the minimum value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}?$ I have tried using the arithmetic mean and harmonic mean inequality but I failed. Not good at inequalities though. Please help.

  • $\begingroup$ Sorry for the bad language. Looking for the minimum value. $\endgroup$
    – reco
    Oct 3, 2017 at 17:49

2 Answers 2


By Cauchy-Schwarz $$(3x+4y+7z)(x^{-1}+y^{-1}+z^{-1})\ge(\sqrt3+\sqrt4+\sqrt7)^2.$$ One can get equality, when the vector $(x,y,z)=t(1/\sqrt3,1/\sqrt4,1/\sqrt7)$ for some $t$. The $t$ in question is $1/(\sqrt3+\sqrt4+\sqrt7)$ and the minimum is $(\sqrt3+\sqrt4+\sqrt7)^2$.

  • $\begingroup$ 12s ahead! +1.. $\endgroup$
    – Macavity
    Oct 3, 2017 at 18:02

From $3x+4y+7z=1$ I got $z= \frac{1}{7} (-3 x-4 y+1)$ and plugged in


gives $$f(x,y)=\frac{7}{1-3 x-4 y}+\frac{1}{x}+\frac{1}{y}$$

$$f'_x=\frac{21}{(1-3 x-4 y)^2}-\frac{1}{x^2};\;f'_y=\frac{28}{(1-3 x-4 y)^2}-\frac{1}{y^2}$$

$f'_x=0$ gives $21x^2=(1-3x-4y)^2$ and $f_y=0\to 28y^2=(1-3x-4y)^2$

so we have $21x^2=28y^2$ which for positive reals means $y=\frac{\sqrt 3}{2}\,x$

furthermore we have from the first equation

$x\sqrt{21}=1-3x-4y$ which after substituting becomes

$x\sqrt{21}=1-3x-2\sqrt{ 3}x$

$$x=\frac{1}{3+2 \sqrt{3}+\sqrt{21}};\;y=\frac{\sqrt{3}}{2 \left(3+2 \sqrt{3}+\sqrt{21}\right)};\;z=\frac{1}{7+2 \sqrt{7}+\sqrt{21}}$$

So minimum of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is

$2 \left(7+2 \sqrt{3}+2 \sqrt{7}+\sqrt{21}\right)\approx 40.676$


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