Ex 1.7(ii) [BOMC-2011]

[rakeen]

$0 \leq \frac{a}{1+b} + \frac{b}{1+a} \leq 1$

Again got the former one but stuck on the later one! $a+b+a^2 +b^2 \geq 0$ and thus we get $0 \leq \frac{a}{1+b} + \frac{b}{1+a}$ need to learn that strategy of solving the later one :'(
-rakeen

[nafistiham]

here the conditions are as follows
$a,b$ are real numbers such that $0\leq a\leq b\leq 1$

[*Mahi*]

Multiply and subtract what you can, and then see what you have to prove $\geq 0$

[nafistiham]

ইশশ!!! এর চেয়ে ছোট সমাধান করতে পারলাম না এটাতেও ভুলভাল আছে কিনা কে জানে

suppose,
\
and,
\[1=a+k+p\]
where,
\[k,p\geq 0\]
now,
\[\frac{a}{1+b}+\frac{b}{1+a}\leq 1\]
\[\Leftrightarrow \frac{a}{1+b}\leq 1-\frac{b}{1+a}\]
\[\Leftrightarrow \frac{a}{1+b}\leq \frac{1+a-b}{1+a}\]
\[\Leftrightarrow \frac{a}{a+k+p+a+k}\leq \frac{a+k+p+a-a-k}{a+k+p+a}\]
\[\Leftrightarrow \frac{a}{2a+2k+p}\leq \frac{a+p}{2a+k+p}\]
\[\Leftrightarrow 0\leq ak+2ap+2kp+p^{2} \]

which is true.

tottal brute force (not good for health)

[*Mahi*]

Okay, here I am!
We have to prove $a^2+a+b+b^2 \leq ab+a+b+1$
Hint:

multiply both sides of $a^2-ab+b^2 \leq 1$ with $a+b$