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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.

mrQWERTY
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    The questions seem either incomplete, or badly translated. – Asaf Karagila Feb 13 '14 at 03:57
  • 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. – mrQWERTY Feb 13 '14 at 04:00
  • Have you proved the Arithmetic Mean Harmonic Mean Inequality? Or AM/GM? – André Nicolas Feb 13 '14 at 04:01
  • You need to remove the "such that." – André Nicolas Feb 13 '14 at 04:01
  • 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 04:01
  • Thanks. Then, the question must not be correctly written. Thank you anyways! – mrQWERTY Feb 13 '14 at 04:03
  • A "trick" for (2) is to look at $(a-b)^2+(b-c)^2+(c-a)^2$. – André Nicolas Feb 13 '14 at 04:04
  • What exactly is the "discrete math" part of this question? – Marc van Leeuwen Feb 13 '14 at 12:36
  • How come no one has mentioned Cauchy-Schwarz for the $1$st ineqality? This is a very obvious application of it. Does that have something to do with the "discrete math" part? I am not very sure what you mean by saying "discrete math". – user26486 Feb 13 '14 at 13:01
  • The last paragraph is irrelevant and makes the question longer. There is no need to add junk to questions just to placate the users opposed to "homework". (1) is the arithmetic mean - harmonic mean inequality for 3 variables. (2) is true without the condition that $a,b,c$ be positive (changing signs does not change the left side but can reduce the right side). Inequality (2) for all real values of $a,b,c$ is the positive-definiteness of a quadratic form, which is equivalent to the ability to write the form as a sum of squares of linear forms, in this case $\sum (a-b)^2$. – zyx Feb 13 '14 at 16:25

4 Answers4

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Hints:

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

Ittay Weiss
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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 $$x+\frac{1}{x}-2=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2.$$

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

lab bhattacharjee
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HINT:

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

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

Bobby
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