Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am stuck on these two problems.

$1$. Prove that for every three positive real numbers a, b, and c that

$(a+b+c)*(\frac{1}{a}+\frac{1}{b} + \frac{1}{c}) \ge 9$.

$2$. Prove that for every three positive real numbers a, b, and c that $a^2 + b^2 + c^2 \ge ab + bc + ac$.

I have tried direct proof and have not gotten anywhere significant. I won't put the work on there since it is way too long and I don't think it will help. There must be some sort of trick involved, but for the life of me, I cannot figure it out.

share|cite|improve this question
The questions seem either incomplete, or badly translated. – Asaf Karagila Feb 13 '14 at 3:57
Sorry, what is confusing about it? I copied the question directly from the original source and it is translated correctly. Please explain further so I can make necessary adjustments. – Someone Feb 13 '14 at 4:00
Have you proved the Arithmetic Mean Harmonic Mean Inequality? Or AM/GM? – André Nicolas Feb 13 '14 at 4:01
You need to remove the "such that." – André Nicolas Feb 13 '14 at 4:01
Froggy, "For every X such that Y" is just giving a condition Y on the object X. You didn't tell us what we need to prove about X. – Asaf Karagila Feb 13 '14 at 4:01

4 Answers 4


1) Without loss of generality, $a\le b\le c$. Then $(a+b+c)\cdot (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})= 3 + \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}$. Which of these ratios is at least $1$?

2) play with $(\pm a \pm b \pm c)^2\ge 0$.

share|cite|improve this answer
Hi, thank you for your answer. I am confused about the ratios. Can you please elaborate further? – Someone Feb 13 '14 at 5:19
Oh I got it. Wow, that is brilliant. – Someone Feb 13 '14 at 5:40

For (2), use the fact that $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0.$$

For (1), use the hint of Ittay Weiss, and the fact that if $x$ is positive, then $x+\frac{1}{x}\ge 2$. This follows from the fact that $$x+\frac{1}{x}-2=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2.$$

share|cite|improve this answer
Or for the less inspired among us, $x+\frac1x\geq2$ (weak inequality!) for $x>0$ follows from study of the extrema of the function $x\mapsto x+\frac1x$. – Marc van Leeuwen Feb 13 '14 at 12:38

For $(1)$ $$a+b+c\ge 3(abc)^{\frac13}$$ and $$\frac1a+\frac1b+\frac1c\ge3 \frac1{(abc)^{\frac13}}$$

share|cite|improve this answer


$(1)$:Use $\frac{(a+b+c)}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

$(2)$: Multiply by $2$: $2a^2 + 2b^2 + 2c^2 -2ab -2bc - 2ac\geq 0$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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