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

Question

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

Answer 1

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.