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Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that

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


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up vote 20 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 $$ a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2 $$

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Note that $$a+b+c≥3, \tag{1}$$ by AM-GM inequality

Now, by Titu's lemma $$a^2+b^2+c^2 ≥\frac{(a+b+c)^2}3 ≥ \frac{(a+b+c)^2}{a+b+c}$$ by using (1)

then $$a^2+b^2+c^2 ≥a+b+c$$ Q.E.D.

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By Titu's lemma you mean this: (Maybe it would be worth adding a link to your post.) – Martin Sleziak Sep 7 '15 at 7:53

We have: $$\sqrt{\frac{a^2+b^2+c^2}{3}}\underset{Q\geq G}{\geq} \sqrt[3]{abc} = 1$$ But $\sqrt{x} \geq 1 \iff x \geq 1 \iff x\geq \sqrt{x}$ so: $$\frac{a^2+b^2+c^2}{3} \geq\sqrt{\frac{a^2+b^2+c^2}{3}}\underset{Q\geq A}{\geq}\frac{a+b+c}{3}$$ $$a^2+b^2+c^2 \geq a+b+c$$

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why if $\sqrt{x} \geq 1$ then the next equivalences? – Iuli Sep 7 '15 at 18:45

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: $$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c $$


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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 $\frac{a+b+c}{3}≥1$

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

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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.

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