Fundamental Inequalities - 3

By Titu's lemma :

$\begin{aligned} \frac {x^2}{x+1} + \frac {y^2}{y+2} & \ge \frac {(x+y)^2}{x+1+y+2} = \boxed{\frac {16}7} & \small \color{#3D99F6} \text{Since }x+y = 4 \end{aligned}$

Equality occurs when $x=\frac 43$ and $y = \frac 83$ .

Substituting for $y$ and simplifying the resulting expression, we get the given expression equal to $\dfrac{x^2-8x-16}{x^2-5x-6}$ . Let this be equal to $a$ . Then $(a-1)x^2+(8-5a) x+(16-6a) =0$ . Since $x$ is real, the discriminant of the equation must be non-negative definite. This condition yields $a\geq \dfrac{16}{7}$ or $a\leq \dfrac{8}{7}$ . Therefore the minimum value of $a$ is $\dfrac{16}{7}$ when $x=\dfrac{24}{18}=\dfrac{4}{3},y=4-\dfrac{4}{3}=\dfrac{8}{3}$

It's given that:

$x+y = 4$ $\dots$ ( EQ 1 )

From the answer choices, we can also see that the value of the equation will be a fraction.

To add fractions, the two fractions need to have the same denominator. So, we can set each denominator above equal to each other, like this:

$x+1 = y+2$ $\dots$ ( EQ 2 )

So now that we have each of these equations, we can solve ( EQ 1 ) for a variable and plug it into ( EQ 2 ) .

Let's solve for $x$ :

$\rightarrow$ $x = 4-y$

$\rightarrow$ $(4-y) + 1 = y+2$

$\rightarrow$ $5-y = y+2$

$\rightarrow$ $3 = 2y$

$\rightarrow$ $y = ^3$ $/_{ 2}$

Now, plugging y into ( EQ 1 ) , we find that $x$ = $^5$ $/_{ 2}$

Plugging the values of x and y that we just solved for into the equation given will give us a possible value for its solution.

$\dfrac{(^5/_2)^2}{(^5/_2)+1}$ + $\dfrac{(^3/_2)^2}{(^3/_2)+2}$ = $\dfrac{25/4}{(^5/_2)+^2/_2}$ + $\dfrac{9/4}{(^3/_2)+(^4/_2)}$

$\rightarrow$ $\dfrac{36/4}{7/2}$ = $\dfrac{9}{7/2}$

$\rightarrow$ $9$ $\times$ $^2/_7$ = $18/7$

So I could be wrong, but I'm pretty sure the correct answer is $18/7$ . But even if not, I figured it was close enough!