Balanced to the minimum

Algebra Level 5

For positive $a,b$ and $c$ , find the minimum value of $\frac{(a+b)^{2}+(a+b+4c)^2}{abc}\, (a+b+c).$

The answer is 100.0.

1 solution

Jessica Wang
Jan 15, 2017

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

By the AM-GM inequality , we have $\begin{aligned} (a+b)^{2}+(a+b+4c)^{2} & =(a+b)^{2}+((a+2c)+(b+2c))^{2} \\ & \geqslant (2\sqrt{ab})^{2}+(2\sqrt{2ac}+2\sqrt{2bc})^{2} \\ & =4ab+8ac+8bc+16c\sqrt{ab}. \end{aligned}$

Therefore,

$\begin{aligned} \frac{(a+b)^{2}+(a+b+4c)^{2}}{abc}\, (a+b+c) &\geqslant \frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}\, (a+b+c) \\ & =\left ( \frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{ab}} \right )\, \left ( a+b+c \right ) \\ & =8\left ( \frac{1}{2c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ab}} \right ) \left ( \frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c \right ) \\ & \geqslant 8\left ( 5\, \sqrt[5]{\frac{1}{2a^{2}b^{2}c}} \right )\left ( 5\, \sqrt[5]{\frac{a^{2}b^{2}c}{2^4}} \right )=100. \end{aligned}$

Thus, the minimum value required is $\boxed{100}$ .

Moderator note:

(Calvin's rant about minimum)

When finding the minimum of a function, we have to show

That the value is a lower bound.
That the value can be achieved.

As pointed out by Mark, this solution only does 1, but not 2.

You have used the AM/GM inequality five times. To achieve the minimum value of $100$ , using your argument, you need to show that all these inequalities can be equalities at the same time (for example when $a=b=2$ , $c=1$ .

Since the quantity you want to minimize is homogeneous in $a,b,c$ , you could add the restriction $a+b+c=1$ without loss of generality, and then you simply need to minimize $\frac{(1-c)^2 + (1 + 3c)^2}{abc} \hspace{2cm} a + b + c = 1\;,\; a,b,c > 0$ One application of the AM/GM inequality will show that this is minimized when $a=b=\tfrac12(1-c)$ , so all you have to do is minimize $\frac{4\big[(1-c)^2 + (1 + 3c)^2]}{c(1-c)^2} \hspace{2cm} 0 < c < 1$ which is minimized when $c=\tfrac15$ .

Mark Hennings - 4 years, 4 months ago

Hmmm. Since the expression is minimized when $a=b\ne c$ , this looks like it can be solved using UVW method . But for some reason I couldn't seem to finish it off...

On the other hand, UVW seems to be more complicated than the standard AM/GM approach that you used. So I guess UVW is not desired here.

Pi Han Goh - 4 years, 4 months ago

Agreed! And I added my usual rant above.

Another approach to finding the equality condition via the existing proof is to track what needs to hold at each stage of inequality. In this proof, we need $a+b = a + 2c = b + 2c$ in the first paragraph, and $2c = b = a = \sqrt{ab}$ , $a = b = 2c$ in the second paragraph. Since we can find a solution set (like $(2, 2, 1)$ ), hence equality can hold througout.

Calvin Lin Staff - 4 years, 4 months ago

It would suit the problem more if it was considered in Algebra.

Achal Jain - 4 years, 4 months ago

I used Titu's Lemma and AM - GM Inequality ...But I got $54*4^{1/3}$

EDITED : I didn't look for the equality case.

Ankit Kumar Jain - 4 years, 4 months ago

Balanced to the minimum

The answer is 100.0.

1 solution

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