If $a+b+c=6$ and $a,b,c$ belongs to positive reals $\mathbb{R}^+$; then find the minimum value of $\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$ .

inequality optimization maxima-minima a.m.-g.m.-inequality cauchy-schwarz-inequality

11,340

Solution 1

We can use also AM-GM.

For $a=1$, $b=2$ and $c=3$ we get a value $6$.

We'll prove that it's a minimal value.

Thus, it's enough to prove that $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\geq6.$$ Indeed, by AM-GM: $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}=a+\frac{1}{a}+b+\frac{4}{b}+c+\frac{9}{c}-6\geq$$ $$\geq2\sqrt{a\cdot\frac{1}{a}}+2\sqrt{b\cdot\frac{4}{b}}+2\sqrt{c\cdot\frac{9}{c}}-6=2+4+6-6=6.$$ Done!

Solution 2

Cauchy-schwarz inequality state that :

For any $x_1, y_1, z_1; x_2, y_2, z_2 \ \in \mathbb{R}^+$; we have: $$ \left(\sqrt{x_1 \cdot x_2} + \sqrt{y_1 \cdot y_2} + \sqrt{z_1 \cdot z_2}\right) ^ 2 \leq \left(x_1 + y_1 + z_1\right) \cdot \left(x_2 + y_2 + z_2\right) \ . $$

So we can conclude that:

$$ \begin{align} & \left( \sqrt{\frac{a}{a}} + \sqrt{\frac{4b}{b}} + \sqrt{\frac{9c}{c}} \right) ^ 2 & \leq \ \ \ \ & \left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right) & \cdot \ \ \ \ \ \ \ \ & (a+b+c) & \Longrightarrow \\ & \ \ \ \ \ \ \ \ \ \ \ \left( 1 + 2 + 3 \right) ^ 2 & \leq \ \ \ \ & \left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right) & \cdot \ \ \ \ \ \ \ \ & \ \ \ \ \ \ \ \ (6) & \Longrightarrow \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( 6 \right) ^ 2 & \leq \ \ \ \ & \left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right) & \cdot \ \ \ \ \ \ \ \ & \ \ \ \ \ \ \ \ (6) & \Longrightarrow \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 & \leq \ \ \ \ & \left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right) \end{align} $$

Note that this in-equlality is sharp for $a=1, b=2, c=3$;
for which one can see the value of $\left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right)$ is equal to $6$.

Solution 3

Using the lagrange multiplier ($k$);

$$f(a,b,c,k)=\frac1a+\frac4b+\frac9c+k(a+b+c-6)$$

We will take the derivative in respect to all the variables;

$$f_k'=0$$

$$f_a'=k-\frac{1}{a^2}$$

$$f_b'=k-\frac{4}{b^2}$$

$$f_c'=k-\frac{9}{c^2}$$

$$k=\frac{1}{a^2}=\frac{4}{b^2}=\frac{9}{c^2}$$

$$\sqrt{k}=\frac1a=\frac2b=\frac3c$$

$$\sqrt{k}=\frac{1+2+3}{a+b+c}=\frac66$$

$$a=1$$ and $$b=2$$ and $$c=3$$ then

$$\frac1a+\frac4b+\frac9c=1+2+3=6$$

Done!!

11,340

Tabber

Updated on August 01, 2022

Comments

Tabber 10 months

If $a+b+c=6$ and $a,b,c$ belongs to positive reals,
then find the minimum value of $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$$

using AM $\ge HM$

$\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

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

or

**why not

$AM\ge GM $

$\frac{\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+a+b+c}{6}\ge (\frac{1}{a}\times\frac{4}{b}\times\frac{9}{c}\times a\times b\times c)^\frac{1}{6} $

$\Rightarrow \frac{1}{a}+\frac{4}{b}+\frac{9}{c}\ge 6(6^\frac{1}{3}-1)$**
Tabber over 5 years

Thank you Famke that answers my question.
Davood over 5 years

@Quintessence ; My dear Quintessence, wellcome to math-stack-exchange .