JEE Inequalities Starter

We first use Titu's Lemma on the given expression that yields us the following inequality:

$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2 \geq \frac{\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{1+1+1}=\frac{\left(6+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}$

Now, to minimize the RHS of this inequality so as to achieve the tightest lower bound, we need to minimize $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ . Now, we'll use the AM-HM inequality on the positive reals $a,b,c$ to obtain:

$\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\\ \implies \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq ~\frac{3}{2}$

and equality occurs at $a=b=c=2$ .

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So, using the minimum value obtained, we can evaluate the original inequality as follows:

$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2 \geq \frac{75}{4}=18.75$

Now, we need to check the equality case to verify that this is indeed the minimum value of the expression. Using the equality case for Titu's Lemma, we can verify that equality occurs when,

$\left(a+\frac{1}{b}\right)=\left(b+\frac{1}{c}\right)=\left(c+\frac{1}{a}\right)=\frac{5}{2}$

We can see that this equality case also occurs at $a=b=c=2$ which was the same for the AM-HM inequality. Thus, the equality cases are matched and the minimum value of the original inequality is confirmed to be $\boxed{18.75}$

We use the inequality $R.M.S \geq A.M$ .

$\sqrt{\dfrac{\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2}{3}} \geq \dfrac{a + b + c + \dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}}{3}$

$\Rightarrow\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \geq \dfrac{\left(a + b + c + \dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\right)^2}{3}\dots (1)$

We also know that :

$(a + b + c)\left(\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\right) \geq 3^2$

$\Rightarrow(6)\left(\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\right) \geq 9$

$\Rightarrow\left(\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\right) \geq \dfrac{9}{6}$

$\Rightarrow\left(\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\right) \geq \dfrac{3}{2} \dots (2)$

Substituting $(2)$ in $(1)$ ,

$\Rightarrow\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \geq \dfrac{\left(6 + \dfrac{3}{2}\right)^2}{3}$

$\Rightarrow\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \geq \dfrac{\left(\dfrac{15}{2}\right)^2}{3}$

$\Rightarrow\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \geq \dfrac{\left(\dfrac{225}{4}\right)}{3}$

$\Rightarrow\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \geq \dfrac{225}{12}$

$\Rightarrow\left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \geq \huge\boxed{18.75}$

Without titu's lemma= If we expand the original expression we get:

$a^2+2\frac{a}{b}+\frac{1}{b^2}+b^2+2\frac{b}{c}+\frac{1}{c^2}+c^2+2\frac{c}{a}+\frac{1}{a^2}$

reordering:

$2*(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+(a^2+b^2+c^2)+(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})$

minimizing each of the three previous terms we get (using AM>=GM inequality):

$2*(3)+3x+3\frac{1}{x}$ (equation z)

$x=(abc)^{2/3}$

using the identity given in the problem we have:

$(a+b+c)^2=6^2$

$a^2+b^2+c^2=36-2(ab+bc+ac)$

if we try to minimize this we get:

$36-2*3x$

so now for the second term we have an easy equation:

$36-6x=3x; x=4$

finally we return to equation z and we get:

$6+3*4+3*(1/4)=18.75$

NOTE: First time using LaTeX :O

Because of symmetry, it is likely that a=b=c is some minimum. With a=b=c=2, the Exp. =18.75. We try a=1, b=2, c=3 and we get Exp. >18.75. So $\boxed{\color{#D61F06}{ \Large18.75 } }$ is the minimum.

We use the RMS-AM followed by the AM-HM inequalities:

$\begin{aligned} \sqrt {\frac{{{{\left( {a + \frac{1}{b}} \right)}^2} + {{\left( {b + \frac{1}{c}} \right)}^2} + {{\left( {c + \frac{1}{a}} \right)}^2}}}{3}} & \ge \frac{{\left( {a + \frac{1}{b}} \right) + \left( {b + \frac{1}{c}} \right) + \left( {c + \frac{1}{a}} \right)}}{3} \;\;\; \mbox{by RMS-AM}\\ & = \frac{{a + b + c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}}{3}\\ & = \frac{6}{3} + \frac{{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}}{3}\\ & \ge 2 + \frac{3}{{a + b + c}} \;\;\; \mbox{by AM-HM}\\ & = 2 + \frac{3}{6} = \frac{5}{2} \end{aligned}$

Squaring both sides and multiplying by 3 yields ${\left( {a + \frac{1}{b}} \right)^2} + {\left( {b + \frac{1}{c}} \right)^2} + {\left( {c + \frac{1}{a}} \right)^2} \ge \frac{{75}}{4} = 18.75$

Use TITU'S LEMMA OR CAUCHY SCWARZ

Use Cauchy-Schwarz inequallity followed by A.M. - H.M. inequality.

Just assume 2 for all of them and do the math. lol