$(a^2+1)(1+1)\ge (a+1)^2$ $(b^2+1)(1+1)\ge (b+1)^2$ $(a^2+1)(b^2+1)\ge (ab+1)^2$ Multiplying these we get that $4(a^2+1)^2(b^2+1)^2\ge \left((a+1)(b+1)(ab+1)\right)^2$ but the relation fromt he hypotese tells us that these are equal. Hence we must have equality in all those 3 inegalities. We get $a=b=1$.

Another solution : we set $a= \tan x, \, b= \tan y$ and we have $\\ 2 = ( \sin x + \cos x ) (\sin y + \cos y ) ( \sin x \sin y + \cos x \cos y ) \Rightarrow \\ \\ 2= 2 \sin \left ( x+ \frac{ \pi}{4} \right ) \sin \left ( y+ \frac{ \pi}{4} \right ) \cos (y-x)$ From this equation if $y,x \in [0,2 \pi )$ we get $y=x= \frac{ \pi }{4}$ or $x =\frac{ \pi }{4} ,y= \pi + \frac{ \pi }{4}$ . Both of them give $a=b=1$ .

another method: $2(a^2+1)(b^2+1)=(a+1)(b+1)(ab+1)$ $\Longleftrightarrow (b^2-b+2)a^2-a(b+1)^2+(2b^2-b+1)=0$ Since $b^2-b+2=(b-\frac{1}2)^2+\frac{7}4>0$ So the discriminant $(b+1)^4-4(b^2-b+2)(2b^2-b+1)=-(b-1)^2(7b^2-2b+7)\leq 0$ . Hence there are only real roots when $b=1$ and this gives us $a=1$

shyong wrote: another method: $2(a^2+1)(b^2+1)=(a+1)(b+1)(ab+1)$ $\Longleftrightarrow (b^2-b+2)a^2-a(b+1)^2+(2b^2-b+1)=0$ Since $b^2-b+2=(b-\frac{1}2)^2+\frac{7}4>0$ So the discriminant $(b+1)^4-4(b^2-b+2)(2b^2-b+1)=-(b-1)^2(7b^2-2b+7)\leq 0$ . Hence there are only real roots when $b=1$ and this gives us $a=1$

kunny wrote: shyong wrote: another method: $2(a^2+1)(b^2+1)=(a+1)(b+1)(ab+1)$ $\Longleftrightarrow (b^2-b+2)a^2-a(b+1)^2+(2b^2-b+1)=0$ Since $b^2-b+2=(b-\frac{1}2)^2+\frac{7}4>0$ So the discriminant $(b+1)^4-4(b^2-b+2)(2b^2-b+1)=-(b-1)^2(7b^2-2b+7)\leq 0$ . Hence there are only real roots when $b=1$ and this gives us $a=1$ anything wrong with that factorization ?? It seems ok to me

Shyong, I think Kunny was puzzled (amased) how u factored this. So am I

delta wrote: Shyong, I think Kunny was puzzled (amased) how u factored this. So am I actually , you can either expand everything out and do the rational roots theorem things . But somehow , from above post , i already know that $b=1$ is the answer . So i guess the discriminant should looks something like $(b-1)^2(pb^2+qb+r)$ . Then you use substitute for some value for left and right and solve simultaneous equation to get $p,q,r$ .

(from C.B.S) (a*a+1)(1+1)>=(a+1)(a+1) (1) equality for a=1 (b*b+1)(1+1)>=(b+1)(b+1) (2) equality for b=1 (a*a+1)(b*b+1)>=(a+b)(a+b) (3) equality for a=b=1 (1)*(2)*(3)→(a*a+1)2*(b*b+1)2 *2*2>=(a+1)2*(b+1)2(a+b)2|√→ (a*a+1)(b*b+1)*2>=(a+1)(b+1)(a+b) equality for a=b=1 In conclusion the solutions of problems are a=b=1

Let $a,b,c$ be reals. Prove that $$(1+a^2)(1+b^2)(1+c^2)\ge(1+ ab)(1+bc )(1+ca )$$ $$\implies$$ Let $a,b$ be real numbers. Prove that $$2(a^2+1)(b^2+1)\ge (a+1)(b+1)(ab+1)$$