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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.