If $x^2+bx+a=0, x^2+ax+b=0$ do not have distinct real roots then find the maximum value of $\frac{a^2+b^2}{a+b}$ if $a,b\gt0$
Solution:
$D\le0\implies b^2-4a\le0, a^2-4b\le0$
It implies $b^2+a^2\le4(a+b)\implies \frac{b^2+a^2}{a+b}\le4$
So, the maximum value is $4$.
Is everything okay with both the question and the answer?
On
This solution looks correct, but the only thing I might add is what @youthdoo alluded to. For what values of $a$ and $b$ do you attain the maximum value of 4? The max is attained when $D=0$ which gives the system of equations: $$(1)\ \ b^2 = 4a$$ $$(2)\ \ a^2 = 4b.$$
Dividing the two equations we obtain: $$\frac{b^2}{a^2} = \frac{a}{b} \implies b^3 = a^3 \implies a^3 - b^3 = 0.$$ Factoring we get $(a-b)(a^2+ab+b^2)=0.$ The only solution is $a=b$ since $a,b>0$. Plugging this back into one of the equations of the system we get $a=b=4$.
As pointed in the comments, I overlooked some details in my previous answer. I am editing it to reflect on some relevant points.
We already know an upper bound for the expression is $4$.
In short, you can represent these inequalities as parabolas in the $1$st quadrant.
$$b^2-4a \le 0$$ $$a^2-4a \le 0$$ $$(a-2)^2+(b-2)^2 = (2\sqrt{2})^2$$
You can interpret the first two inequalities as parabolas and the third as a circle.
To find a maximum, you need to consider points that lie in the region that satisfies the first two inequalities (which is just the region inside the two parabolas), and are closest to the circle described by the third equation.
It is easy to see that in this case the upper bound $4$ is achieved with $(4,4)$.
But if you want distinct equations, you can try to find points with distinct values for the $x$ and $y$ coordinate. Clearly, there is no maximum in this case, but only a strict upper bound of $4$.
If you consider the problem over the distinct integers, this condition is achieved with $(2,1)$ only.