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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.

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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. –  Froggy 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 $(\pm a \pm b \pm c)^2\ge 0$.

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Hi, thank you for your answer. I am confused about the ratios. Can you please elaborate further? –  Froggy Feb 13 '14 at 5:19
Oh I got it. Wow, that is brilliant. –  Froggy 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 $$x+\frac{1}{x}-2=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2.$$

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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 $$\frac1a+\frac1b+\frac1c\ge3 \frac1{(abc)^{\frac13}}$$

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

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

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