# Integrals inequality

I have $$A=\int_1^5{\frac{e^x}{e^x+x^2}dx}$$ $$B=\int_1^5{\frac{x^2}{e^x+x^2}dx}$$ I have already found that $A+B=4$ but now I want to prove that $AB\le4$. I don't know how. I am thinking of using the properties of integrals but nothing seems to work out for me. Any ideas?

• $$\int_1^5{\frac{e^x}{e^x+x^2}dx}+\int_1^5{\frac{x^2}{e^x+x^2}dx}=2$$ $$\int_1^5{\frac{e^x}{e^x+x^2}+\frac{x^2}{e^x+x^2}dx}=2$$ $$\int_1^5{\frac{e^x+x^2}{e^x+x^2}}=2$$ $$\int_1^5{dx}=2$$ $$[x]^5_1=4$$ $$4 \ne 2$$ How? – Yash Jain Jan 30 '18 at 14:01
• sorry misstyped! – Leos Kotrop Jan 30 '18 at 14:06
• It is still $A+B=2$ – Yash Jain Jan 30 '18 at 14:07

This follows from the AM–GM inequality. Both integrals are positive, so $$A+B=4\implies\frac{A+B}2=2\implies \sqrt{AB}\le2\implies AB\le4$$

While Parcly's answer is quick, it can be done without AM-GM as well if you do not happen to know/remember AM-GM. After all, if given $A+B=4$, then the expression $AB$ can be written as $A(4-A)$ which is $-A^2+4A$. The outputs are values according to a parabola that opens down, having vertex $(2,4)$, hence $AB\le4$. Just a thought...

• Which is one way to prove the AM-GM inequality for AM = 4 :) – meh Jan 30 '18 at 22:14

$((\sqrt{A})^2 -(\sqrt{B})^2)^2 \ge 0$
$(\sqrt{A})^2-2\sqrt{AB}+(\sqrt{B})^2 \ge 0$
$(\sqrt{A})^2+(\sqrt{B})^2 \ge 2\sqrt{AB}$
$A + B \ge 2\sqrt{AB}$
so $A + B = 4 \Rightarrow 4 \ge 2\sqrt{AB}$
That is $\sqrt{AB} \le 2$
Therefore $AB \le 4$