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 $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.

Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$


Here's my idea:

$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$

And I'm stuck here.

I need to prove that:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$

In the first case using Cauchy-Schwarz Inequality I prove that:

$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$

$3(a^2 + b^2 + c^2) \ge (a+b+c)^2$

Now I need to prove that:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$

$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$

I need I don't know how to continue.

In the second case I tried proving:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and

$a^2 + b^2 + c^2 \ge a+b+c$

Using Cauchy-Schwarz Inequality I proved:

$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$

$(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$

$a^2 + b^2 + c^2 \ge a+b+c$

But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$

So please help me with this problem.

P.S

My initial idea, which is proving:

$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$

maybe isn't the right way to prove this inequality.

share|improve this question
    
You could try doing something with infinitesimal calculus to show that the minimum of $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2)$ occurs only when there is some other specific relationship between $a$, $b$, and $c$. –  AJMansfield Mar 20 '13 at 23:19

3 Answers 3

up vote 8 down vote accepted

I will use the following lemma (the proof below):

$$2x \geq x^2(3-x^2)\ \ \ \ \text{ for any }\ x \geq 0. \tag{$\clubsuit$}$$

Start by multiplying our inequality by two

$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq 2ab +2bc +2ca, \tag{$\spadesuit$}$$

and observe that

$$2ab + 2bc + 2ca = a(b+c) + b(c+a) + c(b+c) = a(3-a) + b(3-b) + c(3-c)$$

and thus $(\spadesuit)$ is equivalent to

$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq a(3-a) + b(3-b) + c(3-c)$$

which can be obtained by summing up three applications of $(\clubsuit)$ for $x$ equal to $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ respectively:

\begin{align} 2\sqrt{a} &\geq a(3-a), \\ 2\sqrt{b} &\geq b(3-b), \\ 2\sqrt{c} &\geq c(3-c). \\ \end{align}

$$\tag*{$\square$}$$


The lemma

$$2x \geq x^2(3-x^2) \tag{$\clubsuit$}$$

is true for any $x \geq 0$ (and also any $x \leq -2$) because

$$2x - x^2(3-x^2) = (x-1)^2x(x+2)$$

is a polynomial with roots at $0$ and $-2$, a double root at $1$ and a positive coefficient at the largest degree, $x^4$.

$\hspace{60pt}$poly

I hope this helps ;-)

share|improve this answer
    
How do you rearrange? Explain that step, because $$(\sqrt{a} + \sqrt{a})^2 + (2\sqrt{b})^2 + 4c = 4(a+b+c)$$ How do we know that $$a^2 + b^2 + c^2 + 2\sqrt{a} + 2\sqrt{b} + 2\sqrt{c} + 3 \geq (\sqrt{a} + \sqrt{a})^2 + (2\sqrt{b})^2 + 4c$$ ? –  Stefan4024 Mar 21 '13 at 0:04
    
@Stefan4024 That step was wrong, check out the new proof. –  dtldarek Mar 21 '13 at 8:02
1  
Good work!!! Thanks you a lot. It was so good explained that even a kid from first grade will get it. :D –  Stefan4024 Mar 21 '13 at 10:24

From the given inequality $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ observe that $$2(ab+bc+ac)=(a+b+c)^2-a^2-b^2-c^2$$ We can rewrite the original inequality as
$$a^2+2\sqrt{a}+ b^2+2\sqrt{b}+ c^2+2\sqrt{c}\ge9$$ since $(a+b+c)=3$. Using AM-GM set the LHS up as follows: $$a^2+\sqrt{a}+\sqrt{a}\ge3\sqrt[3]{a^2 \sqrt{a}\sqrt{a}}=3a$$ $$b^2+\sqrt{b}+\sqrt{b}\ge3\sqrt[3]{b^2 \sqrt{b}\sqrt{b}}=3b$$ $$c^2+\sqrt{c}+\sqrt{c}\ge3\sqrt[3]{c^2 \sqrt{c}\sqrt{c}}=3c$$ Adding the three inequalities yields $$a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 3(a+b+c) =9 $$ with equality if an only if $a$=$b$=$c$=$1$.

share|improve this answer

Hint:

What lower bound does AM-GM give you when you consider $a^2 + \sqrt{a} + \sqrt{a}$?

Your hope that $\sum \sqrt{a} \ge \sum a = 3$ is false, by using Cauchy Schwarz: $9 = 3(\sum a) \ge (\sum \sqrt{a})^2$. In fact, when $a+b+c = 3$, we have $$\sum a^2 \ge \sum a = 3 \ge \sum \sqrt{a}$$ all by Cauchy-Schwarz, so your hope to split the inequality up is thwarted. This also signals us that we should try to "mix" $a^2$ and $\sqrt{a}$ together in some way, hence the hint.

share|improve this answer

Your Answer

 
discard

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.