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

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.

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 $$ \sum_{cyc}\sqrt{\frac {11(5c + 6a)} {(5a + 6b)(5b + 6c)(5c + 6a)} \cdot a(5b + 6c)} \leq\\ \sqrt{\frac {11^2 (a + b+ c)} {(5a + 6b)(5b + 6c)(5c + 6a)} \sum_{cyc} a(5b + 6c) } =\\ \sqrt{\frac {11^3 (a + b+ c) (ab + bc + ca)} {(5a + 6b)(5b + 6c)(5c + 6a)}} \leq \sqrt 9 = 3 $$