to find the range of $\lambda$

Question

If $a, b, c,d$ are positive real numbers such that $a+b+c+d = 2$, then $\lambda= (1+a+b)(1+c+d)$. then $\lambda$ satisfies? (A) 2$\lt$ $\lambda$$\le3$ (B) 3$\lt$$\lambda$$\le4$ (C)4$\lt$ $\lambda$$\le5$ (D)5$\lt$ $\lambda$$\le6$

what I have tried is $AM \ge GM$ and obtained several equations but I am finding difficulty in proceeding from there.

Answer 1

From the given conditions, you can deduce that $c+d=2-(a+b)$

Now, $\lambda=(1+a+b)(2+(1-(a+b))=(1+(a+b))(3-(a+b))$

Just for making the equation more compact put $a+b=x$ $$\lambda=(1+x)(3-x)=-x^2+2x+3$$

differentiate both sides and equate to zero to obtain the point at which maximum value of the function is obtained.$$-2x+2=0 \implies x=1$$

Plugging this into the equation for $\lambda$, we get the maximum value as $4$

As $0\lt a+b \lt 2$ the lower bound is $3$

Hence $\lambda$ lies between $(3,4]$

NOTE: I felt this method is easier than AM$\ge$GM method so I am including it here.

Answer 2

By AM-GM $$\lambda\leq\left(\frac{1+a+b+1+c+d}{2}\right)^2=4.$$ The equality occurs for $a=b=c=d=\frac{1}{2}$.

Thus, $4$ is a maximum value.

If $a\rightarrow2$ and $b=c=d\rightarrow0$ then $\lambda\rightarrow3$.

We'll prove that it's infimum.

Indeed, we need to prove that $$(1+a+b)(1+c+d)>3$$ or $$(1+a+b)(3-a-b)>3$$ or $$(a+b)(2-a-b)>0$$ or $$(a+b)(c+d)>0.$$ Done!