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.

Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? :


share|cite|improve this question
Also, why do you think it is true? – Thomas Andrews Jul 9 '12 at 18:07
By contradiction ,we can substitute each term in the LHS by x,y,z and we want to prove that $x+y+z \ge1$ but for contradiction we assume $x+y+z<1$,so i think its right – Frank Jul 9 '12 at 19:58
With this reasoning, every statement is false? – Did Jul 10 '12 at 6:14
No,we want to prove something but we assume the opposite to prove what we want by contradiction . – Frank Jul 10 '12 at 11:28
This is not an indication that you tried anything at all. – Did Jul 12 '12 at 19:23

2 Answers 2

up vote 5 down vote accepted

By AM>GM $$ \frac{a+b+c}{3}=1 \Rightarrow abc\le 1 \\ \Rightarrow \frac{1}{2ab^2+1} = \frac{1}{2abc\frac{b}{c}+1}\ge\frac{1}{2\frac{b}{c}+1}=\frac{c}{2b+c} \\ S = \frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge\frac{c}{2b+c}+\frac{a}{2c+a}+\frac{b}{2a+b} $$ $$ \begin{align} 3S-3 & \ge \frac{3c-2b-c}{2b+c}+\frac{3a-2c-a}{2c+a}+\frac{3b-2a-b}{2a+b}\\ & = 2\left(\frac{c-b}{2b+c}+\frac{a-c}{2c+a}+\frac{b-a}{2a+b}\right) \\ & = \frac{2}{D}\left(3ab^2+3bc^2+3ca^2-9abc\right)\\ & = \frac{6abc}{D}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}-3\right) \\ & \ge 0 \end{align} $$ where for clarity we simply write $D$ for the positive denominator, and the last inequality is again by AM>GM $$ 1=\left(\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}\right)^{1/3}\le\frac{1}{3}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right) $$ Finally $3S-3\ge 0 \Rightarrow S\ge 1$.

share|cite|improve this answer

A solution for straightforward mathematicians. :-)

It seems the following.

After multiplication of both sides of the inequality by a common denominator and simplification, we reduce the initial inequality to

$$1+ab^2+bc^2+ca^2\ge 4a^3b^3c^3.$$

Since $$ab^2+bc^2+ca^2\ge 3abc$$ it suffices to check that

$$1+3abc-4(abc)^3\ge 0,$$

that is

$$(1-abc)(2abc+1)^2\ge 0.$$

Tthe last inequality holds because $$abc\le\left(\frac{a+b+c}3\right)^3=1.$$

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.