A technique will solve this easily

Using AM-GM inequality , we have:

$\begin{aligned} a + b + c & \ge 3 \sqrt [3] {abc} \\ 3 & \ge 3 \sqrt [3] {abc} \\ \implies abc & \le 1 \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Equality happens when }a=b=c=1} \\ \color{#3D99F6}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} & \color{#3D99F6}{\ge \frac{3}{\sqrt [3] {(abc)^2}} = 3 \quad \quad \small \text{Equality happens when }a=b=c=1} \\ \color{#D61F06}{a^2 + b^2 + c^2} & \color{#D61F06}{\ge 3 \sqrt [3] {(abc)^2} = 3 \quad \quad \small \text{Equality happens when }a=b=c=1} \end{aligned}$

Therefore,

$\begin{aligned} \color{#3D99F6}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} + \frac{2}{3}(\color{#D61F06}{a^2 + b^2 + c^2}) \ge \color{#3D99F6}{3} + \frac{2}{3}(\color{#D61F06}{3}) = \boxed{5} \quad \quad \small \color{#3D99F6}{\text{Equality happens when }a=b=c=1} \end{aligned}$

Call the expression $P$ , we can rewrite it like this $P=\frac{1}{a^2}+\frac{2}{3}a^2+\frac{1}{b^2}+\frac{2}{3}b^2+\frac{1}{c^2}+\frac{2}{3}c^2$ Now we need to prove that $\frac{1}{a^2}+\frac{2}{3}a^2\geq\frac{5}{3}-\frac{2}{3}(a-1) \ (*)$ $\Leftrightarrow\frac{(a-1)^2(2a^2+6a+3)}{a^2}\geq 0 \ (True \ for\ every \ a>0)$ Same goes for $b$ and $c$ then combine all 3 inequalities $P\geq 5-\frac{2}{3}(a+b+c-3)=5$ So $P_{min}=5$ , the equality holds when $a=b=c=1$

• For any one questioned: how could you arrive at $(*)$ , check out U.C.T (Undefined Coefficient Technique)

Relevant wiki: Arithmetic Mean - Geometric Mean

$\text{The Above Expression can be rewritten as : } \\ (\dfrac1{a}+\dfrac1{b}+\dfrac1{c})^2-2(\dfrac1{ab}+\dfrac1{bc}+\dfrac1{ca})+\dfrac23[(a+b+c)^2-2(ab+bc+ca)] \\ (\dfrac{ab+bc+ca}{abc})^2-2(\dfrac{a+b+c}{abc})+\dfrac23[(a+b+c)^2-2(ab+bc+ca)]$

$\text{It is given that } a+b+c=3 \implies abc\le1 \\ \implies \dfrac{ab+bc+ca}{3}\ge (a^2b^2c^2)^{3} \implies ab+bc+ca\ge 3 \\ \text{Putting the values in the expression, we get :} \\ = 9-6+\dfrac23(9-6) \\ \implies \color{#3D99F6}{\boxed{5}}$

Use TITU'S lemma on (1/a^2 ) +(1/b^2) +(1/c^2) Then by applying am -hm We get the min value of the shown expression as 3 Then we use cauchy Schwarz or titu on a^2 + b^2 +c^2 we get its min value as 3 Then 3×2/3=3 Add 3+2=5

To minimize the equation, let a = b = c = 1. Therefore, $a + b + c = 3$

The equation now simplifies to: $\frac {1} {1^2} + \frac {1} {1^2} + \frac {1} {1^2} + \frac {2(1+1+1)} {3}$ $3 + 2 = 5$ $\boxed{5}$

note that 1/a^2 + 2a^2/ 3 >= 2sqrt(2/3) by AM-GM. By symmetry, note that if we do the same for b and c and add the 3 inequalities then the min value is 3*2sqrt(3) = 4.90 or 5

From titu's lemma, $1/a^2+1/b^2+1/c^2$ is greater than or equal to $9/a^2+b^2+c^2$ Now, by using am-gm on $9/a^2+b^2+c^2$ and $2(a^2+b^2+c^2)/3$ , we get its minimum value as $2\sqrt{6}$ . Can anybody tell me where am I wrong?

(I got this problem right because I entered 4.898)

Using AM-HM,

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

$\implies \frac{1}{a^2} + \frac{1}{b^2}+\frac{1}{c^2} \geq \frac{9}{a^2+b^2+c^2}$

Using Cauchy-Schwarz Inequality ,

$(a^2+b^2+c^2)(1^2+1^2+1^2) \geq (a+b+c)^2 = 9$

$\implies a^2+b^2+c^2 \geq 3$

$\therefore \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2(a^2+b^2+c^2)}{3}$

$\geq \frac{9}{a^2+b^2+c^2} +\frac{2(a^2+b^2+c^2)}{3}$

$\geq \frac{9}{3}+\frac{2(3)}{3}$

$= 3+2$ $=\boxed{5}$ and the equality happens when $a=b=c=1$

The numbers are at minimum when a=b=c=x So x+x+x=3 So x=1 Substitute it into the equation and the minimum is 5