# Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9

Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$

This is how I tried doing it:

Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.

Using AM>GM for $p, q$, I get:

$$\frac{p+q}{2} > {(pq)}^{1/2}$$ $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$ And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$
Thus,
$$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)<9,$$ which is the opposite of what had to be proven. What did I do wrong?

• Your approach shows $\dfrac{a+\frac{1}{a} + b+\frac{1}{b} + c+\frac{1}{c}}{2} \ge 3$, which is correct though not what you wanted to show. Please note that you have not shown that $(a + b +c)( \frac{1}{a} + \frac{1}{b} + \frac{1}{c})<9$. – Macavity Jan 29 '15 at 11:09
• You have shown $\sqrt X \le Y$, and $Y \ge 3$. This does not imply $X < 9$. (BTW, it's $AM \ge GM$, not $AM > GM$.) – TonyK Jan 29 '15 at 11:11
• Okay thanks got it! Btw, if I show that Y < X and Z < Y Then It can be said that Z < X . Right? @TonyK – user165253 Jan 29 '15 at 11:16
• Why is this tagged (linear-algebra). ( I do not see connection to linear algebra.) – Martin Sleziak Jan 29 '15 at 12:44
• @MartinSleziak Sorry, removed it! – user165253 Jan 30 '15 at 7:57

Well, although everyone has answered it, but I think $A.M \geq H.M.$ should also do the work. It's more better in this case. Moreover no one has addressed as to why the OP is wrong.

We know that $\frac{a+b+c}{3} \geq \frac{3}{\frac1a + \frac1b + \frac1c}$.

Henceforth, we get $(a+b+c)(\frac1a + \frac1b + \frac1c) \geq 9$.

Coming to the error in your attempt, in the step where you have written $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$

Clearly, $x+\frac1x \geq 2$, which gives the RHS of the above inequality greater than 3. So therefore, you have said that if $A>B$ and $A>C$ then $B>C$, which is of course not true. Here we cannot compare $B$ and $C$. Hence we couldn't proceed by your method.

• Sorry, didn't read the comments. – Rohinb97 Jan 29 '15 at 14:55
• I actually addressed it in the first version of my answer, but somehow it got lost in later edits. So I added it back to my answer. – user2345215 Jan 29 '15 at 16:08

By the AM-GM inequality, $$\frac{a}{b}+\frac{b}{a}\geq 2$$ and so on, so: $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 3+\sum_{cyc}\left(\frac{b}{a}+\frac{a}{b}\right) \geq 3+6 = \color{red}{9}$$ as wanted.

• What is $\sum_{\text{cyc}}$? – wchargin Jan 29 '15 at 15:11
• @WChargin: $$\sum_{cyc}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b).$$ – Jack D'Aurizio Jan 29 '15 at 15:23

You are using $x+1/x\ge 2$ in the opposite direction. You can do it by AM-GM as follows: $$a+b+c>3\sqrt[3]{abc}\ \ \land\ \ 1/a+1/b+1/c>3/\sqrt[3]{abc}\ \ \implies\ \ (a+b+c)(1/a+1/b+1/c)>9$$

Alternatively, you can use the Cauchy-Schwarz inequality: $$(a+b+c)(1/a+1/b+1/c)>(1+1+1)^2=9$$

Consider $$(a-b)^2 \ge 0$$ $$a^2-2ab+b^2 \ge 0$$ $$a^2+b^2 \ge 2ab$$ $$\frac{a^2+b^2}{ab} \ge 2$$ $$\frac{a}{b}+\frac{b}{a}\ge 2$$ without a loss of generality, we can also say that $$\frac{a}{c}+\frac{c}{a}\ge 2$$ and $$\frac{b}{c}+\frac{c}{b}\ge 2$$ of course $$3=1+1+1=\frac{a}{a}+\frac{b}{b}+\frac{c}{c}$$ So, putting it all together we have $$\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\ge 3+2+2+2$$ $$(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\ge 9$$