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 numbers . Prove the following inequality:

$$\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3.$$

What I tried:

I used Cauchy-Schwarz in the following form $\sqrt{Ax}+\sqrt{By}+\sqrt{Cz} \leq \sqrt{(a+b+c)(x+y+z)}$ for: $$A=11a, \quad{} B=11b, \quad{} C=11c$$ and $$x=\frac{1}{5a+6b}, \quad{} y=\frac{1}{5b+6c}, \quad{} z=\frac{1}{5c+6a}$$ but still nothing. Thanks for your help :)

I tried something else:

$$\large\frac{\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}}}{3} \leq \sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}}$$ and what we have to prove become:

$$\large\sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}} \leq 1 \Leftrightarrow \sqrt{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}} \leq \sqrt{3}$$

Another attempt

$$\large\sqrt{\frac{1}{xy}} \leq \frac{\frac{1}{x}+\frac{1}{y}}{2}=\frac{x+y}{2xy}$$ and $y=1$ and $\displaystyle x=\frac{5a+6b}{11a}$. So:

$$\large\sqrt{\frac{11a}{5a+6b}} \leq \frac{\frac{5a+6b}{11a}+1}{2 \cdot \frac{5a+6b}{11a}}=\frac{8a+3b}{5a+6b}.$$ Now we have to prove: $$\sum_{cyc}{\frac{8a+3b}{5a+6b}} \leq 3.$$

But still nothing .

share|improve this question
2  
Nice question! (+1) –  Chris's sis Sep 10 '12 at 12:26
    
Maybe we have to use the classical method of Lagrange multiplier at last? –  y zhao Sep 10 '12 at 16:22
    
@yzhao Yes, maybe using method of Lagrange the inequality can be proved but this inequality must be done by a children from highschool. –  Iuli Sep 10 '12 at 16:42
add comment

2 Answers

up vote 6 down vote accepted

We can prove the more general inequality for certain values of the ratio between b and a:

$$\sqrt{\frac{(a+b)x}{ax+by}}+\sqrt{\frac{(a+b)y}{ay+bz}}+\sqrt{\frac{(a+b)z}{az+bx}} \leq 3$$ if 0.8152 < b/a < 1.2267

Re-working the LHS and applying Cauchy-Schwarz we get:

$$(\sqrt{az + bx}\sqrt{\frac{(a + b) x}{(a z + b x) (a x + b y)}} + \sqrt{ax + by}\sqrt{\frac{(a + b) y}{(a x + b y) (a y + b z)}} + \sqrt{ay + bz}\sqrt{\frac{(a + b) z}{(ay + bz) (a z + b x)}})^2 \leq (a y + b z + a z + b x + a x + b y)(\frac{(a + b) x}{(a z + b x) (a x + b y)}+\frac{(a + b) y}{(a x + b y) (a y + b z)}+\frac{(a + b) z}{(ay + bz) (a z + b x)}) =\frac{(a + b)^3 (x + y + z) (y z + x y + xz)}{(a x + b y) (b x + a z) (a y + b z)} $$ We are now looking for an upper bound of this last expression. Denote this upper bound by k and consider the expression: $$G=(a + b)^3 (x + y + z) (y z + x y + zx) - k (a x + b y) (b x + a z) (a y + b z)$$ This needs to be always negative if k is an upper bound. Simplifying it turns out that we need a value of k such that:

$$G=((a+b)^3-a^2 b k)(y^2 z+x z^2+x^2 y)+((a+b)^3-a b^2 k)(xy^2+yz^2+zx^2)+(3 (a+b)^3-a^3 k-b^3 k)xyz<=0$$

Applying AM/GM we get:

$$(a^2 b k-(a+b)^3)(y^2 z+x z^2+x^2 y)+(a b^2 k-(a+b)^3)(xy^2+yz^2+zx^2)>=3((a^2 b k-(a+b)^3)+(a b^2 k-(a+b)^3))xyz$$

In order for this to hold we need to assume that (A1) $$a^2 b k-(a+b)^3>=0$$ and (A2) $$a b^2 k-(a+b)^3>=0$$

So now we can choose k so that the coefficients before xyz in the last two expressions are equal. In other words we need k such that:

$$3(a + b)^3 - a^3 k - b^3 k = 3 ((a^2 b k - (a + b)^3) + (a b^2 k - (a + b)^3))$$

It is easy to see that k=9 we completes the proof if our assumptions (A1) and (A2) hold. They hold for a small range of b/a ratios - approximately 0.8152 < b/a < 1.2267.

In order for A1 and A2 to hold for k=9 we can rewrite them in the form $$\frac{(a + b)^3}{a^2 b}<=9 \quad and \quad \frac{(a + b)^3}{a b^2}<=9$$

Letting x=b/a this means that we need the positive values of x for which both:$$ 1/x + 3 x + x^2 - 6<=0 \quad and \quad 1/x^2 + 3/x + x - 6<=0$$

Solving the resulting cubics we get:$$ 2 - \sqrt{3} Cos(\pi/18) + 3 Sin(\pi/18)\le\frac{b}{a}\le-1 + 3 Cos(\pi/9) - \sqrt{3} Sin(\pi/9) $$ or numerically 0.8152 < b/a < 1.2267. This is the range of possible valus for b/a for which our inequality always holds.

You can also note that the product of the interval ends is equal to 1.

share|improve this answer
    
I don't know if it is ok . I don't manage anything . It must be a trick. –  Iuli Sep 10 '12 at 14:49
    
Try this: imgur.com/WSBXh –  ivan Sep 10 '12 at 15:24
    
Can you held how you applied $AM-GM$. Till there I understand what you have done. Thanks :) –  Iuli Sep 11 '12 at 7:09
    
$$zy^2+xz^2+yx^2>=3xyz$$ –  ivan Sep 11 '12 at 9:28
    
I don't understand the transition from : $$(3(a+b)^3-a^3k-b^3k)xyz$$ to $$3((a^2bk-(a+b)^3)+(ab^2k-(a+b)^3))xyz$$ thnaks:) –  Iuli Sep 11 '12 at 9:35
show 8 more comments

It's a bit ugly (actually more than a bit), but it works.

By AM-GM we get $$ abc = \sqrt[3]{ab^2\cdot bc^2 \cdot ca^2} \leq \frac 1 3 (ab^2 + bc^2 + ca^2) $$ and therefore $$ 9(5a + 6b)(5b + 6c)(5c + 6a) - 11^3(a + b + c)(ab + bc + ca) =\\ 289(ab^2 + bc^2 + ca^2) + 19(a^2b + b^2c + c^2a) - 924abc =\\ 289(ab^2 + bc^2 + ca^2 - 3abc) + 19(a^2b + b^2c + c^2a - 3abc) \geq 0 $$ From the previous inequality and applying Cauchy-Schwarz we arrive to $$ \sum_{cyc}\sqrt{\frac {11(5c + 6a)} {(5a + 6b)(5b + 6c)(5c + 6a)} \cdot a(5b + 6c)} \leq\\ \sqrt{\frac {11^2 (a + b+ c)} {(5a + 6b)(5b + 6c)(5c + 6a)} \sum_{cyc} a(5b + 6c) } =\\ \sqrt{\frac {11^3 (a + b+ c) (ab + bc + ca)} {(5a + 6b)(5b + 6c)(5c + 6a)}} \leq \sqrt 9 = 3 $$

share|improve this answer
add comment

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.