Take the 2-minute tour ×
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 $a,b,c$ be positive real numbers such that $abc=1$. Prove that

$a^2+b^2+c^2 \geq a+b+c$.


share|improve this question
add comment

5 Answers

up vote 16 down vote accepted

Using Cauchy-Schwarz inequality we get $$ a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1} $$ From AM-GM we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so $$ \sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2} $$ From $(1)$ and $(2)$ it follows $$ a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2 $$

share|improve this answer
@Graphth The question is not tagged homework....? –  Martin Sleziak Oct 8 '12 at 13:47
@MartinSleziak Neither are the questions that Norbert is downvoting because complete solutions were given. math.stackexchange.com/a/208341/8671 –  Graphth Oct 8 '12 at 13:58
@Graphth I think pointing Norbert to the meta thread where this is discussed would be a more efficient way of communication. I hope that you are aware of the fact that serial donwvoting is discouraged. –  Martin Sleziak Oct 8 '12 at 14:05
@Grapth You act like 4 year old kid. Don't you think there is a difference in difficulty of finding pattern in 1, -3, -7 and proving inequalities. Anyway I appreciate your work of digging into my answers and attempts to find violation of my complete solution principle. This confirm the first statement in this comment - you act like a 4-years old kid –  Norbert Oct 8 '12 at 14:14
add comment

Let's solve it in an elementary way and start from the fact that: $$a^2 \ge 2a -1 \tag1$$ $$b^2 \ge 2b-1 \tag2$$ $$c^2 \ge 2c-1 \tag3$$

Then add up $(1)$ $(2)$ $(3)$ and get: $$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag4$$ By AM-GM we have $$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1 $$ $$ a+b+c \ge 3 \tag5 $$ Finally, from $(4)$ and $(5)$ we obtain the required inequality: $$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c $$


share|improve this answer
add comment

We will use the following form of Cauchy-Schwarz inequality:

From Cauchy-Schwarz inequality applied on the vectors $\displaystyle{ \left( \frac{x_1}{\sqrt{a_1}} , \frac{x_2}{\sqrt{a_2}} , \cdots , \frac{x_n}{\sqrt{a_n}} \right)}$ and $ \displaystyle{ \left( \sqrt{a_1} ,\sqrt{a_2} , \cdots , \sqrt{a_n} \right) }$ where $ x_1 ,x_2 \cdots ,x_n \in \mathbb R $ and $ a_1, a_2, \cdots ,a_n >0 $ we get that

$ \displaystyle{ \frac{x_1^{2}}{a_1} +\frac{x_2^{2}}{a_2} + \cdots + \frac{x_n^{2}}{a_n} \geq \frac{\left(x_1 + x_2 + \cdots + x_n \right)^{2}}{a_1 +a_2 + \cdots + a_n} }$

Back to our problem now the given inequality can be written equivelant in the form

$ \displaystyle{ \frac{a^2}{abc} + \frac{b^2}{abc} +\frac{c^2}{abc} \geq a+ b+c \quad (\star)}$

Using now the inequality we stated in the begging we get that the left-hand-side of $(\star)$ is :

$ \displaystyle{ \frac{a^2}{abc} + \frac{b^2}{abc} +\frac{c^2}{abc} \geq \frac{ \left(a+b+c \right)^2}{3abc} = \frac{ \left(a+b+c \right)^2}{3} \geq \left(a+b+c \right) \cdot \frac{ 3 \sqrt[3]{abc}}{3} = a+b+c }$

which is what we need to prove. Q.E.D

P.S. I the last step we use the AM-GM inequality: $ \displaystyle{ a+b+c \geq 3 \sqrt[3]{abc} }$.

P.S The above form of Cauchy-Schwarz inequality is called Cauchy-Schwarz in Engel form.

share|improve this answer
add comment

As A.M. of any set of positive number ≥ G.M,

$\frac{a+b+c}{3}≥ (abc)^{\frac{1}{3}}$

Now, $ 3(a^2+b^2+c^2)-(a+b+c)^2= (a-b)^2+(b-c)^2+(c-a)^2≥ 0$

=>$ a^2+b^2+c^2≥ \frac{(a+b+c)^2}{3}$

=>$ a^2+b^2+c^2≥(a+b+c)(\frac{a+b+c}{3})≥a+b+c$ as $\frac{a+b+c}{3}≥1$

share|improve this answer
add comment

For a more geometric proof:

Let $C$ be a cuboid whose edges are $a$, $b$ and $c$. If $\text{diam}(C)=\sqrt{a^2+b^2+c^2}$ is fixed, then $a+b+c$ is at most $\sqrt{3}\text{diam}(C)$. So $a+b+c \leq \sqrt{3(a^2+b^2+c^2)}$.

Then, if $\text{diam}(C)$ is fixed, $\text{Vol}(C)=abc$ is at most $\displaystyle \left( \frac{a^2+b^2+c^2}{3} \right)^{3/2}$ (you maximize the volume of a cuboid inscribed into a sphere of radius $\sqrt{a^2+b^2+c^2}/2$). So $a^2+b^2+c^2 \geq 3$.

Finally, $a+b+c \leq a^2+b^2+c^2$.

share|improve this answer
add comment

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.