Find the global maximum of $\frac{\sqrt {xy}}{(x + y + 2)^2},$ where $x,y\geq 0 $

Question

I tried AM-GM inequality and got some conclusions then came to the max value = 1/16 where $x = y = 1$ (still not sure 100%). But it was more likely "guessing" method. We can analyse the function and etc. But if you can then, please, show me more than 1 method. Cause it's school level olympiad problem which doesn't actually consider high school knowledge, so no calculus. I would like to see how to solve it like a student from a middle school, but other methods are also approved. Thanks

Answer 1

Use the 4-term AM-GM to show that

$$ \frac{ x + y + 1 + 1 } { 4 } \geq \sqrt[4]{x\times y \times 1 \times 1 }.$$

Hence, $$ \frac{\sqrt{ xy} }{(x+y+2)^2} = \left( \frac{\sqrt[4]{xy} } { ( x+y+2)} \right)^2 \leq \left( \frac{1}{4} \right)^2 = \frac{1}{16}.$$

Equality holds iff $ x = y = 1 = 1$.

Answer 2

As you've already tried, we can use the inequality of arithmetic and geometric means to get

$$\frac{\sqrt{xy}}{(x+y+2)^2} \le \frac{x+y}{2(x+y+2)^2} \tag{1}\label{eq1A}$$

For simpler algebra, let $z = x + y + 2$ (as suggested in lone student's comment, versus my original $z = x +y$). Then, with the RHS of \eqref{eq1A}, we have for some non-negative $k$ that

$$\begin{equation}\begin{aligned} \frac{z-2}{2z^2} & = k \\ z - 2 & = 2kz^2 \\ (2k)z^2 - z + 2 & = 0 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Using the quadratic formula, for \eqref{eq2A} to have any real roots, the discriminant must be non-negative, i.e.,

$$d = 1 - 4(2k)(2) \ge 0 \;\;\to\;\; 16k \le 1 \;\;\to\;\; k \le \frac{1}{16} \tag{3}\label{eq3A}$$

This shows the maximum possible value of the expression being asked about (i.e., the LHS of \eqref{eq1A}) is $\frac{1}{16}$. Since, as you've already determined, this occurs for $x = y = 1$, this means that is actually the expression's maximum value.

Answer 3

Using $A.M. \geq G.M.$, we have $$ \begin{aligned} \frac{\sqrt{x y}}{(x+y+2)^2} & \leqslant \frac{x+y}{2(x+y+2)^2} \\ & =\frac{x+y+2}{2(x+y+2)^2}-\frac{1}{(x+y+2)^2} \\ & =\frac{1}{2(x+y+2)}-\frac{1}{(x+y+2)^2}=E \end{aligned} $$

So $\frac{\sqrt{x y}}{(x+y+2)^2}$ attains its maximum value $E_{max}$ when and only when $x=y$.

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Now let $k=\frac{1}{x+y+2}$, then \begin{aligned} E & =\frac{1}{2} k-k^2 \\ & =-\left(k-\frac{1}{4}\right)^2+\frac{1}{16} \\ & \leqslant \frac{1}{16} \end{aligned}

Hence $$E_{\text {max }}=\frac{1}{16} \textrm{ at }k=\frac{1}{4} . $$

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Now we can conclude that $\frac{\sqrt{x y}}{(x+y+2)^2}$ attains its maximum value $M$

$$\boxed{M=E_{\text {max }}=\frac{1}{16}} $$ $$(\Leftrightarrow x=y \textrm{ and }\frac{1}{x+y+2}=k=\frac{1}{4} \Leftrightarrow x=y=1.)$$

Answer 4

The solutions you get using the AM-GM inequality are very fast. In this answer, we'll go to the solution just using various substitutions :

Substituting $\thinspace x=(m+n)^2,\thinspace y=(m-n)^2$ then we have :

$$u(m,n):=\frac {|m^2-n^2|}{4(m^2+n^2+1)^2}$$

Again substituting $\thinspace m^2=a,\thinspace n^2=b$ where $a≥b≥0$, yields :

$$v(a,b):=\frac {a-b}{4(a+b+1)^2}$$

Finally substituting $\thinspace a=r+s,\thinspace b=r-s$, where $r≥s≥0$, then :

$$ \begin{align}w(r,s):&=\frac {s}{2(2r+1)^2}\\ &≤\frac {r}{2(2r+1)^2}\end{align} $$

This leads to :

$$ \begin{align}&8wr^2+r(8w-1)+2w=0\\ \implies &\Delta_r=1-16w≥0\\ \implies &w≤\frac {1}{16}\end{align} $$

where $w\equiv w(r,r)\thinspace .$

Equality holds iff, when $\Delta_r=0$ or $r=s=\frac 12$, which corresponds to $x=y=1\thinspace .$

Answer 5

You want to show that for $x,y\geq 0,$

$$ {1 \over 16}\geq {\sqrt{xy} \over (x+y+2)^2},$$

or equivalently, $$ x+y+2\geq 4(xy)^{1/4}.\quad (1)$$

Note that since $z\to \log z$ is concave, Jensen's inequality tells us

$$\small \log(x+y+2)=\log \left(4\left(x+y+1+1\over 4\right)\right)=\log 4+\log \left(x+y+1+1\over 4\right)\geq \log 4+{1 \over 4}\log x+{1 \over 4}\log y,$$

and applying the antilog to both sides gives you $(1).$