Practice #5

Solution:

For $\large\boxed{a,b,c>0}$

By Titu's Lemma ,

$\begin{aligned}\large\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)&\ge \large \dfrac{(1+1+1)^{2}}{a+b+b+c+c+a}\end{aligned}$

Multiplying by $a+b+c$ ,

$\begin{aligned}\large(a+b+c)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)&\ge \large\cancel{(a+b+c)}\times \dfrac{(1+1+1)^{2}}{2\cancel{(a+b+c)}}\\ &\ge\dfrac{9}{2}\\ &=4.5\end{aligned}$

Minimum Value: $\boxed{4.5}$

By AM-HM inequality :

$\begin{aligned} \frac 3{\dfrac 1{a+b}+\dfrac 1{b+c}+\dfrac 1{c+a}} & \le \frac {a+b+b+c+c+a}3 \\ \frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a} & \ge \frac 9{2(a+b+c)} \\ \implies (a+b+c) \left(\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}\right) & \ge \frac 92 = \boxed{4.5} \end{aligned}$

$(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})=\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{b+c}{b+c}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}+3$

$\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}\geq 1.5$ (by Nesbitt's inequality)

so the minimum value of $(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$ is $3+1.5=4.5$