By applying AM-GM inequality, we have $\dfrac{2}{a+1}+ \dfrac{a+1}{2} \ge 2$. Therefore $a+2b+ \frac{2}{a+1} \ge \frac a 2 + 2b+ \frac 32$. By AM-GM again we have $\frac a 2 + \frac b 2 \ge 1$ then $\frac a 2+2b+ \frac 32 \ge \frac 32 b+ \frac 52$. Similarly, we also have $b+2a+ \dfrac{2}{b+1} \ge \frac 52+ \frac 32 a$. Hence $ \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geq \left( \frac 52+ \frac 32 a \right) \left( \frac 52+ \frac 32 b \right)$. Finally, $\left( \frac 52+ \frac 32 a \right) \left( \frac 52+ \frac 32 b \right)= \frac{25}{4}+ \frac{15}{4}(a+b)+ \frac 9 4ab \ge 16$ since $ab \ge 1$. The equality holds if and only if $a=b=1$.

Igor wrote: Show that \[\left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right)\geq 16\] for all positive real numbers $a$ and $b$ such that $ab\geq 1$. By Cauchy Schwarz \begin{align*}\left(a+b+b+\dfrac{2}{a+1}\right)\left(b+a+a+\dfrac{2}{b+1}\right) & \ge \left(a+b+\sqrt{ab}+\frac2{\sqrt{(a+1)(b+1)}}\right)^2 \\ & \ge \left(a+b+\sqrt{ab}+\frac2{\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})}\right)^2\end{align*} Note that $a+b \ge \frac12(\sqrt{a}+\sqrt{b})^2 \ge 2$ Hence by AM-GM we get \[2\frac{(\sqrt{a}+\sqrt{b})^2}4+\sqrt{ab}+\frac2{\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})} \ge 4\sqrt[4]{\sqrt[4]{ab}\cdot\frac{(\sqrt{a}+\sqrt{b})^3}8} \ge 4\] which proves the inequality. Equality holds iff $a=b=1$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3109858 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2787943

\[ \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geq 16 \] $ \Leftrightarrow $ $2a^2 + 2b^2 + 5ab +$ $\dfrac{2a+4b}{b+1}+$ $\dfrac{2b+4a}{a+1}+ $ $\dfrac {4}{(a+1)(b+1)}$ $\ge 16$ $ \Leftrightarrow$ $ 2a^2 + 2b^2 + 5ab +$ $\dfrac{2a+4b}{b+1}+$ $\dfrac{2b+4a}{a+1}+ $ $\dfrac {4}{(a+1)(b+1)}$ $\ge$ $9ab +$ $\dfrac{2a+4b}{b+1}+$ $\dfrac{2b+4a}{a+1}+ $ $\dfrac {4}{(a+1)(b+1)}$ $\ge$ $9 +$ $\dfrac{2a+4b}{b+1}+$ $\dfrac{2b+4a}{a+1}+ $ $\dfrac {4}{(a+1)(b+1)}$ $\ge 16$ $ \Leftrightarrow$ $\dfrac{2a+4b}{b+1}+$ $\dfrac{2b+4a}{a+1}+ $ $\dfrac {4}{(a+1)(b+1)}$ $\ge 7$ $ \Leftrightarrow$ $ 2a^2 + 2b^2 + 6a + 6b + 8ab + 4 \ge 7ab + 7a + 7b +7$ $ \Leftrightarrow$ $ 2a^2 + 2b^2 + ab \ge a + b + 3$ $ \Leftrightarrow$ $ 2a^2 + 2b^2 \ge a + b + 2$ $ \Leftrightarrow$ $ a^2 + b^2 \ge a + b $ Now, assuming that if $ab \ge 1$ at least one from $a,b$ is $ \ge 1$, wlog let $a \ge 1$ and $a=b+x$, $x \ge 0$. From this substitutions, we get $b^2 +bx \ge 1$ and $b+x \ge 1$ $ a^2 + b^2 \ge a + b $ $ \Leftrightarrow$ $ \Leftrightarrow$ $ 2b^2 +2bx + x^2 \ge 2b + x$ $ \Leftrightarrow$ $ 2b^2 +2bx + x^2 \ge b^2 + bx +1 + x^2 \ge 2b + x$ $ \Leftrightarrow$ $ (b-1)^2 + bx + x^2 \ge x$ $ \Leftrightarrow$ $ (b-1)^2 + x(b+x-1) \ge 0$ which is true. Equality holds when $a=b=1$.

$LHS=\left(\frac{a+b}{2}+\frac{a+b}{2}+b+\frac{2}{a+1}\right)\left(\frac{a+b}{2}+\frac{a+b}{2}+a+\frac{2}{b+1}\right)\ge$ $16\sqrt[4]{\frac{(a+b)^4ab}{4(a+1)(b+1)}}\ge 16\sqrt[4]{\frac{(a+b)^2\cdot4(ab)^2}{4(a+1)(b+1)}}\ge$ $16\sqrt[4]{\frac{a^2+b^2+2ab}{a+b+ab+1}}\ge16\sqrt[4]{\frac{\sqrt{ab}(a+b)+ab+1}{a+b+ab+1}}\ge 16$

The following inequality is also true. For all non-negatives $a$ and $b$ prove that: \[ \left(a+2b+\frac{12}{a+1}\right)\left(b+2a+\frac{12}{b+1}\right)\geq 81 \]

arqady wrote: The following inequality is also true. For all non-negatives $a$ and $b$ prove that: \[ \left(a+2b+\frac{12}{a+1}\right)\left(b+2a+\frac{12}{b+1}\right)\geq 81 \] Rewrite: $ \Big( 2b+ \frac{a^2+a+12}{a+1}\Big) \Big( 2a+ \frac{b^2+b+12}{b+1}\Big) \ge 81$ Because: $ \frac{a^2+a+12}{a+1} =\frac{(a^2+1)+a+11}{a+1} \ge \frac{3a+11}{a+1} = \frac{8}{a+1}+3$ and a similary I.E we need to prove: $ \Big( 2b+\frac{8}{a+1}+3\Big) \Big( 2a+ \frac{8}{b+1}+3\Big) \ge 81$ $ \Leftrightarrow \Big( 2y+ \frac{8}{x}+1\Big) \Big( 2x+\frac{8}{y}+1\Big) \ge 81$ (1) With: $x=a+1$ and $y=b+1$ Use Cauchy – Schwarz Inequality we have: LHS(1) $ \ge \Big( 2\sqrt{xy}+ \frac{8}{\sqrt{xy}}+1\Big)^2 \ge (2*4+1)^2 =81$

When does the equality hold perfect_square ??

The original inequality can be solved like this: We write each term as $(a+2b+ \frac{2}{a+1}) \ge(a+2b+\frac{2}{a(b+1)}) =(a+\frac{b+1}{2}+\frac{2}{a(b+1)}+\frac{3b-1}{2})$ $ \ge 3+\frac{3b-1}{2} =\frac{3b+5}{2} \ge 4.b^{\frac{3}{8}} $ So similarly treating the other term as same we get the whole expression $\ge 16.(ab)^{\frac{3}{8}} \ge 16$. Hence proved.

perfect_square wrote: Because: $ \frac{a^2+a+12}{a+1} =\frac{(a^2+1)+a+11}{a+1} \ge \frac{3a+11}{a+1} ...$ Equality holds $ \Rightarrow a=b=1$ And if $a=b=1$ then Equality holds

arqady wrote: The following inequality is also true. For all non-negatives $a$ and $b$ prove that: \[ \left(a+2b+\frac{12}{a+1}\right)\left(b+2a+\frac{12}{b+1}\right)\geq 81 \] $LHS= \left(a+1+\frac{4}{a+1}+2(b+1)+\frac{8}{a+1}-3\right)\left(b+1+\frac{4}{b+1}+2(a+1)+\frac{8}{b+1}-3\right)\ge$ $ \left(1+8\sqrt{\frac{b+1}{a+1}}\right)\left(1+8\sqrt{\frac{a+1}{b+1}}\right)\ge(1+8)^2=81$

Since $ ab\ge\ 1 $ we have $ a+b\ge\ 2, \sqrt{ab}\ge\ 1 $. Now, we have $ (a+b+b+\frac{2}{a+1})(a+b+a+\frac{2}{b+1})\ge\ (2+b+\frac{2}{a+1})(2+a+\frac{2}{b+1})=(b+1+1+\frac{2}{a+1})(a+1+1+\frac{2}{b+1})\ge\ (\sqrt{(a+1)(b+1)}+1+\frac{2}{\sqrt{(a+1)(b+1}})^2 $. So, it's enough to prove that $ \sqrt{(a+1)(b+1)}+\frac{2}{\sqrt{(a+1)(b+1)}}\ge\ 3 $. Let $ \sqrt{(a+1)(b+1)}=x $. $ x+\frac{2}{x}\ge\ 3 $ or $ (x-1)(x-2)\ge\ 0 $. Now we have to prove that $ \sqrt{(a+1)(b+1)}\ge\ 2, (a+1)(b+1) \ge\ 4 $ which is true since $ a+b\ge\ 2 $ and $ ab\ge\ 1 $

Generalization of JBMO 2013 Problem 3 https://jbmo2013.tubitak.gov.tr/Problems.html https://jbmo2013.tubitak.gov.tr/sites/default/files/jbmo2013solutions.pdf

By Wieghted AM-GM, $ (a+2b+\frac{2}{a+1}) =\frac{a^2 + a + 2ab + 2b + 2}{a+1} =\frac{(a^2 + a + \frac{3}{2}ab + \frac{1}{2}b) + (\frac{1}{2}ab + \frac{3}{2}b + 1 + 1)}{a+1} \geq \frac{4\sqrt[4]{a^{9/2}b^2} + 4\sqrt[4]{a^{1/2}b^2}}{a+1} = \frac{4a^{9/8}b^{1/} + 4a^{1/8}b^{1/2}}{a+1} = 4a^{1/8}b^{1/2}$ Thus $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1})\geq 16a^{5/8}b^{5/8} \geq 16$

note that , $a+2b+\frac{2}{a+1} =\frac{a+1}{2}+\frac{a+1}{2} +\frac{2}{a+1} +2b-1$ so , by AM-GM , $a+2b+\frac{2}{a+1} \ge \frac{3+a+4b}{2}$ so , $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1}) \ge \frac{25+15(a+b)+9ab}{4} \ge 16$

Igor wrote: Show that \[\left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right)\geq 16\] for all positive real numbers $a$ and $b$ such that $ab\geq 1$. Since , \[ a+1+\frac{4}{a+1} \ge 4 \] adding $ 4b+a $ to both sides gives \[ a+2b+\frac{2}{a+1} \ge \frac{3+a+4b}{2} \] Similarly one can get \[ b+2a+\frac{2}{b+1} \ge \frac{3+b+4a}{2} \] Thus , multiplying gives \[ \left(a+2b+\frac{2}{a+1}\right)\left( b+2a+\frac{2}{b+1}\right) \ge \frac{3+a+4b}{2} \cdot \frac{3+b+4a}{2} \] Now by cauchy swartz \[ \frac{3+a+4b}{2} \cdot \frac{3+b+4a}{2} \ge \frac{(3+\sqrt{ab}+4\sqrt{ab})^2}{4} \ge \frac{64}{4} \ge 16 \Box \]

E(a,b)=(a+b+2/(a+1) )*(b+2a+2/(b+1)) a+2/(a+1)>=(a+3)/2|*2*(a+1) (1) 2*a*a+2*a+4>=a*a+4*a+3|-3-a*a-4*a  a*a-2*a+1>=0  (a-1)(a-1)>=0 b+2/(b+1)>=(b+3)/2 (2) From (1)+(2)=>E(a,b)>=((a+3)/2+2*b)((b+3)/2+2*a)=(a+4*b+3)/2*(b+4*a+3)/2>=(√a*b)+4*√(a*b)+3) *(√(a*b)+4*√(a*b)+3)/4(from C.B.S)>=64/4=16 Equality for a=b=1. In conclusion the solution of the problem is a=b=1.

Solution: We have, $a+2b+\frac {2}{a+1}=2b+\frac {a^2+a+2}{a+1}=1+2b+\frac {a^2+1}{a+1} $. Also, $\frac {a^2+1}{a+1}\ge \frac {a+1}{2} $. Thus, $(a+2b+\frac {2}{a+1})(b+2a+\frac {2}{b+1})\ge (\frac {a+4b+3}{2})(\frac {b+4a+3}{2})$. But, $(\frac {a+4b+3}{2})(\frac {b+4a+3}{2})=\frac {17ab+\sum_{a,b} {(4a^2+15a)} +9}{4}\ge \frac {17+8+30+9}{4}=16$.

We will use that $a+b \geq 2$, so: $a+2b+\frac{2}{a+1} \geq b+2+\frac{2}{a+1}=\frac {b+1}{2}+\frac{b+1}{2}+\frac {2}{a+1}+1$ By $AM-GM$ inequality: $\frac{b+1}{2}+\frac{b+1}{2}+\frac{2}{a+1}+1 \geq 4\sqrt[4]{\frac{(b+1)^2}{2(a+1)}}$ $b+2a+\frac{2}{b+1} \geq a+2+\frac{2}{b+1}=\frac {a+1}{2}+\frac{a+1}{2}+\frac {2}{b+1}+1$ By $AM-GM$ inequality: $\frac{a+1}{2}+\frac{a+1}{2}+\frac{2}{b+1}+1 \geq 4\sqrt[4]{\frac{(a+1)^2}{2(b+1)}}$ Now, using $AM-GM: (a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1}) \geq 16\sqrt[4]{\frac{(a+1)(b+1)}{4}} \geq 16\sqrt[8]{ab} \geq 16$ Equality is when $a=b=1$. That is all .

Solution from Twitch Solves ISL: It's enough to prove \[ \left( (b+1) + \frac{2}{a+1} + 1 \right) \left( \frac{2}{b+1} + (a+1) + 1 \right) \ge 16. \]Let $x = a+1$ and $y = b+1$, so $xy \ge ab+a+b+1 \ge ab+2\sqrt{ab}+1=4$. Then \begin{align*} \left( y+\frac2x+1 \right)\left( \frac2y+x+1 \right) &= \frac{4}{xy} + xy + 1 + 2 + 2 + x + y + \frac 2x + \frac 2y \\ &= 5 + \frac{4}{xy} + xy + x + y + \frac 2x + \frac 2y \\ &= 5 + 4 \cdot \frac{1}{xy} + xy + x + y + 2 \cdot \frac 1x + 2 \cdot \frac 1y \\ &\ge 5 + 4 \cdot \frac{1}{xy} + xy + 2\left( \sqrt{xy} + \frac{2}{\sqrt{xy}} \right) \end{align*}On the other hand, it's easy to prove the following two assertions by clearing the denominator: For any $t \ge 4$, we have $t + \frac 4t \ge 5$. For any $t \ge 2$, we have $t + \frac 2t \ge 3$. Applying this finishes the problem.

Just a bunch of separate bounds. After expanding, $$ab+2a^2+2b^2+\frac{2a}{b+1}+\frac{2b}{a+1}+4ab+\frac{4b}{b+1}+\frac{4a}{a+1}+\frac{4}{(a+1)(b+1)}\geq 16.$$Now we just do the following bounds: $5ab\geq 5$, $2a^2+2b^2\geq 4$, $\frac{4a+4}{a+1}+\frac{4b+4}{b+1}\geq 8$ and $$\frac{2a}{b+1}+\frac{2b}{a+1}+\frac{4}{(a+1)(b+1)} -\frac{4}{a+1}-\frac{4}{b+1}\geq -1.$$After summation of those inequalies we obtain desired inequality. The first three are obvious, thus we only show that the last is also true. \begin{align*} \frac{2a}{b+1}+\frac{2b}{a+1}+\frac{4}{(a+1)(b+1)} -\frac{4}{a+1}-\frac{4}{b+1}&\geq -1 \Longleftrightarrow \\ 2a(a+1)+2b(b+1)+4&\geq -ab-a-b-1+4(b+1)+4(a+1) \Longleftrightarrow\\ 2a^2+2b^2&\geq a+b+3-ab \Longleftrightarrow \\ 2a^2+2b^2&\geq a^{\frac{3}{2}}b^{\frac{1}{2}}+a^{\frac{1}{2}}b^{\frac{3}{2}}+2ab, \end{align*}which is true by AM-GM.

shinichiman wrote: By applying AM-GM inequality, we have $\dfrac{2}{a+1}+ \dfrac{a+1}{2} \ge 2$. Therefore $a+2b+ \frac{2}{a+1} \ge \frac a 2 + 2b+ \frac 32$. By AM-GM again we have $\frac a 2 + \frac b 2 \ge 1$ then $\frac a 2+2b+ \frac 32 \ge \frac 32 b+ \frac 52$. Similarly, we also have $b+2a+ \dfrac{2}{b+1} \ge \frac 52+ \frac 32 a$. Hence $ \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geq \left( \frac 52+ \frac 32 a \right) \left( \frac 52+ \frac 32 b \right)$. Finally, $\left( \frac 52+ \frac 32 a \right) \left( \frac 52+ \frac 32 b \right)= \frac{25}{4}+ \frac{15}{4}(a+b)+ \frac 9 4ab \ge 16$ since $ab \ge 1$. The equality holds if and only if $a=b=1$. How did you get therefore $a+2b+ \frac{2}{a+1} \ge \frac a 2 + 2b+ \frac 32?$

$\textbf{\color{red}{A different approach with AM-GM:}}$ We have \begin{align*} \left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right)&\geqslant \left(a+\frac{1+b}{2}+\frac{2}{a(1+b)}+\frac{3}{2}b-\frac{1}{2}\right)\left(b+\frac{1+a}{2}+\frac{2}{b(1+a)}+\frac{3}{2}a-\frac{1}{2}\right) \\ &\geqslant \left(3+\frac{3}{2}b-\frac{1}{2}\right)\left(3+\frac{3}{2}a-\frac{1}{2}\right) \\&= \frac{1}{4}(3b+5)(3a+5) \\ &\geqslant \frac{1}{4}(6\sqrt b+2)(6\sqrt a+2) \\ &=9\sqrt{ab}+3(\sqrt a +\sqrt b)+1 \\ &\geqslant 9+6\sqrt[4]{ab} +1 \\ &\geqslant 9+6 +1 \\ &=16. \quad \blacksquare \end{align*}

math31415926535 wrote: shinichiman wrote: By applying AM-GM inequality, we have $\dfrac{2}{a+1}+ \dfrac{a+1}{2} \ge 2$. Therefore $a+2b+ \frac{2}{a+1} \ge \frac a 2 + 2b+ \frac 32$. By AM-GM again we have $\frac a 2 + \frac b 2 \ge 1$ then $\frac a 2+2b+ \frac 32 \ge \frac 32 b+ \frac 52$. Similarly, we also have $b+2a+ \dfrac{2}{b+1} \ge \frac 52+ \frac 32 a$. Hence $ \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\geq \left( \frac 52+ \frac 32 a \right) \left( \frac 52+ \frac 32 b \right)$. Finally, $\left( \frac 52+ \frac 32 a \right) \left( \frac 52+ \frac 32 b \right)= \frac{25}{4}+ \frac{15}{4}(a+b)+ \frac 9 4ab \ge 16$ since $ab \ge 1$. The equality holds if and only if $a=b=1$. How did you get therefore $a+2b+ \frac{2}{a+1} \ge \frac a 2 + 2b+ \frac 32?$ anyone?

Since $x+\frac2{x+1}\ge\frac x2+\frac32$. $\Leftrightarrow x+\frac4{x+1}\ge3$ $\Leftrightarrow x(x+1)+4-3(x+1)\ge0$ $\Leftrightarrow(x-1)^2\ge0$

Let $a , b$ be positive real numbers such that $ab\geq 1$.Show that $$\left(a+2b+\dfrac{2}{b+1}\right)\left(b+2a+\dfrac{2}{a+1}\right)\geq 16$$Solution: $$a+2b+\dfrac{2}{b+1}=\dfrac{b+1}{2}+\dfrac{2}{b+1}+a+\dfrac{3b}{2}-\dfrac{1}{2}\geq\dfrac{2a+3b+3}{2}$$$$\left(a+2b+\dfrac{2}{b+1}\right)\left(b+2a+\dfrac{2}{a+1}\right)\geq \dfrac{2a+3b+3}{2}\cdot \dfrac{3a+2b+3}{2}$$$$=\dfrac{1}{4}(6(a^2+b^2)+13ab+15(a+b)+9)\geq 16$$ Let $a , b$ be positive real numbers such that $ab\geq 1$. Show that for $\lambda\ge \dfrac{1}{2}$ we have \[\left(a+\lambda b+\dfrac{2}{b+1}\right)\left(b+\lambda a+\dfrac{2}{a+1}\right)\geq (\lambda+2)^2.\]Let $a , b,c$ be positive real numbers such that $abc\geq 1$. Show that \[\left(a+2b+\dfrac{2}{b+1}\right)\left(b+2c+\dfrac{2}{c+1}\right)\left(c+2a+\dfrac{2}{a+1}\right)\geq 64 .\]Let $a , b, c$ be positive real numbers such that $abc\geq 1$. Show that for $\lambda\ge \dfrac{1}{2}$ we have \[\left(a+\lambda b+\dfrac{2}{b+1}\right)\left(b+\lambda c+\dfrac{2}{c+1}\right)\left(c+\lambda a+\dfrac{2}{a+1}\right)\geq (\lambda+2)^3 .\]Let $x_1,x_2, \cdots, x_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n\geq 1$. Prove that $$\left(a_1+2a_2+\frac{2}{a_3+1} \right) \left(a_2+2a_3+\frac{2}{a_4+1} \right)\cdots \left(a_{n-1}+2a_n+\frac{2}{a_1+1} \right)\left(a_n+2a_1+\frac{2}{a_2+1} \right)\geq 4^n$$Let $x_1,x_2, \cdots, x_n>0 (n\ge 2)$ and $a_1a_2\cdots a_n\geq 1$. Prove that $$\left(a_1+2a_2+\frac{2}{a_2+1} \right) \left(a_2+2a_3+\frac{2}{a_3+1} \right)\cdots \left(a_{n-1}+2a_n+\frac{2}{a_n+1} \right)\left(a_n+2a_1+\frac{2}{a_1+1} \right)\geq 4^n$$h h h h Let $a, b >0$ and $ab(a+1)(b+1)=(ab+1)^2$. Prove that $$ \left(a+2b+\frac{1}{a} \right) \left(b+2a+\frac{1}{b} \right) \geq 16$$Let $a, b >0$ and $ab(a+1)(b+1)=(ab+1)^2$. Prove that$$ \left(a+2b+\frac{1}{b} \right) \left(b+2a+\frac{1}{a} \right) \geq 16$$

sqing wrote: Let $a, b >0$ and $ab(a+1)(b+1)=(ab+1)^2$. Prove that $$ \left(a+2b+\frac{1}{b} \right) \left(b+2a+\frac{1}{a} \right) \geq 16$$ But my problem is claim that $ab \leq 1$ ! It is so interesting.

Very nice problem! Here is my solution , same as in #17 - posting for storage

Let $s=a+b$, $p=ab$. Clearing the denominators gives $((a+2b)(a+1)+2)((b+2a)(b+1)+2) \geq 16(a+1)(b+1)$. The right-hand side equals $16(s+p+1)$, the left-hand one is $$ (a^2+a+2ab+2b+2)(b^2+b+2ab+2a+2) = (a^2+b+2p+s+2)(b^2+a+2p+s+2)$$$$ = (a^2+b)(b^2+a) + (a^2+b^2+a+b)(2p+s+2) + (2p+s+2)^2 $$$$ = p^2 + p + s^3 - 3sp + (s^2-2p+s)(2p+s+2) + (2p+s+2)^2 $$$$ = p^2 + (2s^2 + s + 5)p + 2s^3 + 4s^2 + 6s + 4 $$and so now it suffices to prove $$ p^2 + (2s^2 + s - 11)p + s(2s^2 + 4s - 10) - 12 \geq 0. $$We are given $p\geq 1$ and so $s \geq 2\sqrt{p} \geq 2$, whence $2s^2 + s - 11 \geq -1$ and $s(2s^2+4s-10) \geq 12$. So the left-hand side of the latter is at least $p(p-1) \geq 0$. Equality holds iff $p = 1$ and $s=2\sqrt{p}$, i.e. only for $a=b=1$.

Note that $a+b \ge 2\sqrt{ab} \ge 2$ and $(a^2+b^2) \ge 2ab \ge 2$. Have no idea so just expand: $(a+2b+\frac{2}{a+1})(b+2a+\frac{2}{b+1}) = ab + 2a^2 + \frac{2a}{b+1} + 2b^2 + 4ab + \frac{4b}{b+1} + \frac{2a}{a+1} + \frac{4a}{a+1} + \frac{4}{(a+1)(b+1)} = (ab + 2(a+b)^2) + \frac{2a^2+2a^2+6a+6b+8ab+4}{(a+1)(b+1)} \ge 9 + \frac{2a^2+2a^2+6a+6b+8ab+4}{(a+1)(b+1)}$ so we need to prove $2a^2+2a^2+6a+6b+8ab+4 \ge 7ab+ 7a + 7b + 7$ or $2a^2 + 2b^2 + ab + 4 \ge a + b + 7$ or $2a^2 + 2b^2 \ge a + b + 2$ or $a^2 + (a^2+1) + b^2 + (b^2+1) \ge a + b + 4$ or $a^2 + a + b^2 + b \ge 4$ which is true.

Beautiful

mihaig wrote: Beautiful indeed,good ideas! this it is one old idea from Marius Stanean.

Let $a , b$ be positive real numbers such that $ab\geq 1$.Show that $$\left(a^2+2b+\dfrac{2}{a+1}\right)\left(b^2+2a+\dfrac{2}{b+1}\right) \geq 16$$$$ \left(a+2b+\dfrac{2}{a^2+1}\right)\left(b+2a+\dfrac{2}{b^2+1}\right) \geq 16$$

Show that : \[\left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right)\leq 16.\] for all positive real numbers $a$ and $b$ such that : $a+b\leq 2$.

Let $a , b$ be positive real numbers such that $a^2+b^2\leq 2$.Show that $$\left(a+2b+\dfrac{2}{a+1}\right)\left(b+2a+\dfrac{2}{b+1}\right) \leq 16$$$$\left(a+2b+\dfrac{2}{b+1}\right)\left(b+2a+\dfrac{2}{a+1}\right) \leq 16$$Let $a , b$ be positive real numbers such that $ab\geq 1$.Show that

The original problem is a special case of the following generalizations (#48,49,40) where $$\phi=1,\lambda=2,k=1$$.

Generalization 1 Let $a,b$ be positive reals and $\lambda$ be integer ($\lambda \geq 1$) such that $ab\geq 1$. Then prove that $$\left(a+\lambda b+\dfrac{\lambda }{a+1}\right)\left(b+\lambda a+\dfrac{\lambda }{b+1}\right)\geq \left(\dfrac{\left(\lambda +1\right)\left(\lambda +2\right)-4}{\lambda }\right)^2$$

Generalization 2 Let $a,b,k$ be positive reals and $\lambda$ be integer ($\lambda \geq 1$) such that $ab\geq k$. Then prove that $$\left(a+\lambda b+\dfrac{\lambda }{a+k}\right)\left(b+\lambda a+\dfrac{\lambda }{b+k}\right)\geq \left(\dfrac{\left(\lambda +1\right)\left(\lambda +2\right)}{\lambda }\right)^2.\sqrt[\lambda ^2+3\lambda -2]{k^{\left(\lambda ^2-1\right)}}$$

Generalization 3 Let $a,b,k$ be positive reals and $\lambda,\phi$ be integers ($\lambda,\phi \geq 1$) such that $ab\geq k$. Then prove that $$\left(\phi a+\lambda b+\dfrac{\lambda }{a+k}\right)\left(\phi b+\lambda a+\dfrac{\lambda }{b+k}\right)\geq \left(\lambda ^2+\lambda\left(\phi +2\right)-2\right)^2\left(\sqrt[\lambda ^2+\lambda \left(\phi+2\right)-2]{k^{\left(\lambda ^2+\lambda \left(\phi -1\right)-1\right)}}\right)$$

\[\Pi(a+2b+\frac{a+1}{2})\geq \Pi(b+\frac{a+1}{2}+\frac{a+1}{2}+\frac{2}{a+1})\geq \Pi{ (b+3\sqrt[3]{\frac{a+1}{2}})}\geq (a+3)(b+3)\geq (ab+9)+6\sqrt{ab}\geq 16\]As desired.$\blacksquare$