$a^2+b^2\ge2ab , a^2+b^2\ge2 , 2ab\ge2 $son iki eşitsizliği taraf tarafa toplarsak;
$(a+b)^2\ge4 , a+b\ge2 , \dfrac{a}{2}+\dfrac{b}{2}\ge1$ olur;
$\dfrac{2}{a+1},\dfrac{a+1}{2}$ sayılarına A.G.O. uygularsak;
$\dfrac{2}{a+1}+\dfrac{a+1}{2}\ge 2$ buradan $\dfrac{2}{a+1}\ge\dfrac{3-a}{2}=\dfrac{3}{2}-\dfrac{a}{2}$ olur;
Buradan da $ a+2b+\dfrac{2}{a+1}\ge\dfrac{a}{2}+2b+\dfrac{3}{2} $ olur;
$\dfrac{a}{2}+\dfrac{b}{2}\ge1$ olduğundan $\dfrac{a}{2}\ge1-\dfrac{b}{2}$ olur;
Buradan da $ \dfrac{a}{2}+2b+\dfrac{3}{2}\ge\dfrac{3}{2}b+\dfrac{5}{2} $ olur;
Aynısını diğer bölüm için uygularsak $ b+2a+\dfrac{2}{b+1}\ge\dfrac{5}{2}+\dfrac{3}{2}a $ elde ederiz;
Sonuc olarak $ \left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right)\geq\left(\dfrac{5}{2}+\dfrac{3}{2}a\right)\left(\dfrac{5}{2}+\dfrac{3}{2}b\right) $ elde ederiz;
$ \left(\dfrac{5}{2}+\dfrac{3}{2}a\right)\left(\dfrac{5}{2}+\dfrac{3}{2}b\right)=\dfrac{25}{4}+\dfrac{15}{4}(a+b)+\dfrac{9}{4}ab$ ve $ a+b\ge2 , ab\ge1 $ durumlarını göz önüne alırsak ;
$ \left(\dfrac{5}{2}+\dfrac{3}{2}a\right)\left(\dfrac{5}{2}+\dfrac{3}{2}b\right)=\dfrac{25}{4}+\dfrac{15}{4}(a+b)+\dfrac{9}{4}ab\ge 16 $ ispat burada biter;
$ \left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right)\ge\left(\dfrac{5}{2}+\dfrac{3}{2}a\right)\left(\dfrac{5}{2}+\dfrac{3}{2}b\right)=\dfrac{25}{4}+\dfrac{15}{4}(a+b)+\dfrac{9}{4}ab\ge 16 $ olur .
