# Min and Max of:$\sqrt{ab}+\frac{5}{4}\sqrt{bc}+\sqrt{ca}$

Let: $$a,b,c\geq 0$$ such that $$a+b+c=25$$. Find the Min and Max value of: $$\sqrt{ab}+\frac{5}{4}\sqrt{bc}+\sqrt{ca}$$

For the minimum, I don't know how to do that. The problem is from this original problem which I have changed the variable:

$$x,y,z$$ are real numbers such that: $$9x^2+4y^2+4z^2=25$$. Find the Min Max value of: $$6xy+5yz+6zx$$

For the maximum value, I tried this:

Suppose when the maximum value happened: $$x=my=nz$$

So get this: $$6xy+5yz+6zx = \frac{1}{m}x(my)+\frac{1}{mn}(my)(nz)+\frac{1}{n}(nz) \leq (\frac{1}{2m}+\frac{1}{2n})x^2+ (\frac{m}{2}+\frac{m}{2n})y^2+(\frac{n}{2}+\frac{n}{2m})z^2$$

Then from this, we have to find $$m,n$$ such that:

$$\frac{\frac{1}{2m}+\frac{1}{2n}}{9}=\frac{\frac{m}{2}+\frac{m}{2n}}{4}=\frac{\frac{n}{2}+\frac{n}{2m}}{4}$$

which is unsolvable

For a maximal value smoothing helps.

We need to find a minimal value of $$k$$, for which the inequality $$ab+\frac{5}{4}bc+ca\leq k\frac{a^2+b^2+c^2}{25}$$ is true for any positives $$a$$, $$b$$ and $$c$$.

It's obvious that such $$k_{min}$$ is positive.

Let $$b+c=constant$$.

Thus, we obtain: $$a(b+c)+\frac{5}{4}bc+\frac{2kbc}{25}\leq\frac{k(b+c)^2}{25},$$ which says that it's enough to solve our problem for a maximal value of $$bc$$, which happens for $$b=c$$.

Let $$a=xb$$.

Thus, we need to find a minimal value of $$k$$, for which the inequality $$\frac{k}{25}x^2-2x+\frac{2k}{25}-\frac{5}{4}\geq0$$ is true for any positive value of $$x$$, which gives $$1-\frac{k}{25}\left(\frac{2k}{25}-\frac{5}{4}\right)\leq0$$ or $$k\geq\frac{125+75\sqrt{17}}{16}.$$ The equality occurs for $$b=c$$ and $$a=\frac{25}{k}b$$, where $$a^2+b^2+c^2=25$$ and $$k=\frac{125+75\sqrt{17}}{16},$$ which says that $$\frac{125+75\sqrt{17}}{16}$$ is a maximal value.