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?
Best Answer
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 $a=b=4$.