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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 .

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    $\begingroup$ Nice question! (+1) $\endgroup$ Sep 10, 2012 at 12:26
  • $\begingroup$ Maybe we have to use the classical method of Lagrange multiplier at last? $\endgroup$
    – zy_
    Sep 10, 2012 at 16:22
  • $\begingroup$ @yzhao Yes, maybe using method of Lagrange the inequality can be proved but this inequality must be done by a children from highschool. $\endgroup$
    – Iuli
    Sep 10, 2012 at 16:42

2 Answers 2

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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.

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  • $\begingroup$ I don't know if it is ok . I don't manage anything . It must be a trick. $\endgroup$
    – Iuli
    Sep 10, 2012 at 14:49
  • $\begingroup$ Try this: imgur.com/WSBXh $\endgroup$
    – ivan
    Sep 10, 2012 at 15:24
  • $\begingroup$ Can you held how you applied $AM-GM$. Till there I understand what you have done. Thanks :) $\endgroup$
    – Iuli
    Sep 11, 2012 at 7:09
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    $\begingroup$ $$zy^2+xz^2+yx^2>=3xyz$$ $\endgroup$
    – ivan
    Sep 11, 2012 at 9:28
  • $\begingroup$ 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:) $\endgroup$
    – Iuli
    Sep 11, 2012 at 9:35
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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 $$ \sum_{cyc}\sqrt{\frac {11(5c + 6a)} {(5a + 6b)(5b + 6c)(5c + 6a)} \cdot a(5b + 6c)} \leq\\ \sqrt{\frac {11^2 (a + b+ c)} {(5a + 6b)(5b + 6c)(5c + 6a)} \sum_{cyc} a(5b + 6c) } =\\ \sqrt{\frac {11^3 (a + b+ c) (ab + bc + ca)} {(5a + 6b)(5b + 6c)(5c + 6a)}} \leq \sqrt 9 = 3 $$

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