Minimum of $\frac{1}x + \frac{10}y$, for $x,y>0$ such that $x+y+\sqrt{xy} =3$?

Question

In this question, I have tried many different methods and algebraic equations but I can't find a way to get to the answer.

Let $x$ and $y$ be positive real numbers such that $x+y+\sqrt{xy} =3$ Then min value of $\frac{1}x + \frac{10}y$ is ?

Ans is given 49/6

I took $x$ and $y$ to RHS and squared and some algebraic changes to reach $(x-y)^2 + 3(x-2)(y-2) = 3$

But am stuck at this point. I thought maybe I could get some help from Engels inequality but that also took me nowhere.

Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. — Martin R, Dec 11 '23 at 09:21
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At the very least, say which chapter of your lectures is concerned by these exercises. Are you studying conic curves, extremas with Lagrange multipliers, etc. ? — Jean Marie, Dec 11 '23 at 09:43
Yes I am studying conic sections but not so sure about the other 2 — Yash Shrivastava, Dec 11 '23 at 10:04
Therefore $49/6$ is the maximal value* of the expression in the left hand side. — Jean Marie, Dec 11 '23 at 16:38
Explanation: if, whatever the value of $x$ and $y$, we have $1/x+10/y \le 49/8$ it means that $49/8$ is larger of equal to the maximal value that $1/x+10/y$ can reach. — Jean Marie, Dec 11 '23 at 17:21
According to Engels inequality $\frac {x1^2}{a1} + \frac {x2^2}{a2} + ........... +\frac {xn^2}{an} ≤ \frac {(x1+ x2+ .......... + xn)^2}{a1+a2+............+an}$ — Yash Shrivastava, Dec 12 '23 at 11:38

score 4 · Accepted Answer · answered Dec 11 '23 at 23:31

Let $k = \sqrt{\frac{y}{x}} > 0$. We have $y = k^2 x$. The given condition is written as $x + k^2 x + kx = 3$ which results in $$x = \frac{3}{k^2 + k + 1}, \quad y = \frac{3k^2}{k^2 + k + 1}.$$

We need to find the minimum of $$f(k) := \frac{k^2 + k + 1}{3} + \frac{10(k^2 + k + 1)}{3k^2}.$$

We have $$f'(k) = \frac{(k - 2)(2k^3 + 5k^2 + 10k + 10)}{3k^3}.$$

Thus, $f(k)$ is strictly decreasing on $(0, 2)$, and strictly increasing on $(2, \infty)$. The minimum of $f(k)$ is achieved at $k = 2$. We have $f(2) = 49/6$.

The minimum is $49/6$.

score 2 · Answer 2 · answered Dec 11 '23 at 20:33

This is not a proof, just a commentary on your attempt and a visual representation.

I am puzzled by this exercise which necessitates indeed to have a certain knowledge of conic curves (you say you have this knowledge) but also of techniques of minimization that look beyond the use of simple differentiation (and you say you don't know the technique of Lagrange multipliers).

Nevertheless, I show you what I have done graphically without saying it is a solution.

Fig. 1: Hyperbolas $(C'_m)$ for $m$ ranging from $30/6$ (the highest) to $49/6$, the lowest, which is the solution, just tangent to the arc of ellipse.

First of all, your transformation by squaring into :

$$(x-y)^2 + 3(x-2)(y-2) = 3$$

brings in a "spurious arc" of curve ; indeed, to the (blue) curve with equation

$$(C) \ : \ \ \ \ x+y+\sqrt{xy}=3,$$

you add the (cyan) curve $x+y-\sqrt{xy}=3$ which is "parasitic".

How are the red curves defined ? By parametric equations :

$$(C'_m) \ : \ \ \ \ \frac{1}{x}+\frac{10}{y}=m$$

which are branches of hyperbolas.

The question boils down to find the largest value of $m$ for which there is a point $(x,y)$ belonging to $C$ and $C'_m$. Indeed, this looks to happen for $m=49/6$. But a visual conviction is not a proof.

SAGE program which has generated Fig. 1 :

var('x y');
g=implicit_plot(x+y+sqrt(x*y)-3, (x,0,5), (y,0,5),color='blue');
h=implicit_plot(x+y-sqrt(x*y)-3, (x,0,5), (y,0,5),color='cyan');
g=g+h;
for m in range(30,49) :
h=implicit_plot((10/y+1/x-m/6),(x,0,5),(y,0,5),color='red');
g=g+h;
show(g);

Minimum of $\frac{1}x + \frac{10}y$, for $x,y>0$ such that $x+y+\sqrt{xy} =3$?

2 Answers2