1. ## inequality

hello,i tried to solve the following inequality but found difficulties i tried AM GM but didn't seem to work

a,b,c are positive real numbers with a+b^²+c^4=28
prove that abc≤64
any help or advice is highly appreciated

by amgm i had 28/3≥cubicroot(ab^2c^4) but didn't know how to advance further i did some other attempts but i failed

2. ## Re: inequality

I am not sure that this helps because it uses calculus.

$d > 0,\ e > 0\, f > 0, \text { and } d + e + f = 28.$

$z = def - y(28 - d - e - f) \implies$

$\dfrac{\delta z}{\delta d} = ef - y \implies ef = y \text { if } \dfrac{\delta z}{\delta d} = 0.$

$\dfrac{\delta z}{\delta e} = df - y \implies df = y \text { if } \dfrac{\delta z}{\delta e} = 0.$

$\dfrac{\delta z}{\delta f} = de - y \implies de = y \text { if } \dfrac{\delta z}{\delta f} = 0.$

$\dfrac{\delta z}{\delta y} = 28 - d - e - f \implies d + e + f = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

$ef = y \implies y > 0 \text { and } f = \dfrac{y}{e}.$

$df = y \implies d * \dfrac{y}{e} = y \implies \dfrac{d}{e} = 1 \implies d = e.$

$de = y \text { and } f = \dfrac{y}{e} \implies f = \dfrac{de}{e} = d = e \implies$

$d = e = f = \dfrac{28}{3}.$

In other words, to maximize (d * e * f) subject to (d + e + f) = 28, d = e = f = 28/3. How you discover that with algebra, I am not sure, but I suppose you can confirm it with algebra.

$\therefore a = \dfrac{28}{3} < 9.5,\ b = \sqrt{\dfrac{28}{3}} < 3.1 , \text { and } c = \sqrt[4]{\dfrac{28}{3}} < 1.8.$

$\therefore abc < 9.5 * 3.1 * 1.8 = 53.01 \le 64.$

3. ## Re: inequality

Originally Posted by JeffM
I am not sure that this helps because it uses calculus.

$d > 0,\ e > 0\, f > 0, \text { and } d + e + f = 28.$

$z = def - y(28 - d - e - f) \implies$

$\dfrac{\delta z}{\delta d} = ef - y \implies ef = y \text { if } \dfrac{\delta z}{\delta d} = 0.$

$\dfrac{\delta z}{\delta e} = df - y \implies df = y \text { if } \dfrac{\delta z}{\delta e} = 0.$

$\dfrac{\delta z}{\delta f} = de - y \implies de = y \text { if } \dfrac{\delta z}{\delta f} = 0.$

$\dfrac{\delta z}{\delta y} = 28 - d - e - f \implies d + e + f = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

$ef = y \implies y > 0 \text { and } f = \dfrac{y}{e}.$

$df = y \implies d * \dfrac{y}{e} = y \implies \dfrac{d}{e} = 1 \implies d = e.$

$de = y \text { and } f = \dfrac{y}{e} \implies f = \dfrac{de}{e} = d = e \implies$

$d = e = f = \dfrac{28}{3}.$

In other words, to maximize (d * e * f) subject to (d + e + f) = 28, d = e = f = 28/3. How you discover that with algebra, I am not sure, but I suppose you can confirm it with algebra.

$\therefore a = \dfrac{28}{3} < 9.5,\ b = \sqrt{\dfrac{28}{3}} < 3.1 , \text { and } c = \sqrt[4]{\dfrac{28}{3}} < 1.8.$

$\therefore abc < 9.5 * 3.1 * 1.8 = 53.01 \le 64.$

There is something wrong with the last part of this argument.

if $\displaystyle a=16, b=2\sqrt{2}, c=\sqrt{2}$ then $\displaystyle a b c=64$ giving the maximum of $\displaystyle a b c$

so $\displaystyle a b c$ cannot be less than 53.01

4. ## Re: inequality

You are correct. I think I better give it up.

I need to be looking at $d + e^2 + f^4= 28$ as the constraint if dealing with $z = def.$

I'll work on that.

5. ## Re: inequality

i found abc≤28²/8c because

(a+b^2+c^4)^2≥4a(b^2+c^4)≥8abc^2

6. ## Re: inequality

$a > 0,\ b > 0\, c > 0, \text { and } a + b^2 + e^4 = 28.$

$z = abc - y(28 - a - b^2 - c^2) \implies$

$\dfrac{\delta z}{\delta a} = bc - y \implies bc = y \text { if } \dfrac{\delta z}{\delta a} = 0.$

$\dfrac{\delta z}{\delta b} = ac - 2by \implies ac = 2by \text { if } \dfrac{\delta z}{\delta b} = 0.$

$\dfrac{\delta z}{\delta c} = ab - 4c^3y \implies ab = 4c^3y \text { if } \dfrac{\delta z}{\delta c} = 0.$

$\dfrac{\delta z}{\delta y} = 28 - a - b^2 - c^4 \implies a + b^2 + c^4 = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

$bc = y \implies y > 0 \text { and } c = \dfrac{y}{b}.$

$ac = 2by \implies a * \dfrac{y}{b} = 2by \implies a = 2b^2.$

$ab = 4c^3y \implies 2b^3 = 4c^3 * bc \implies 0.5b^2 = c^4 \implies c = \sqrt[4]{0.5b^2}.$

$\therefore 28 = a + b^2 + c^4 = 2b^2 + b^2 + 0.5b^2 = 3.5b^2 \implies$

$0.5 * 7b^2 = 28 \implies 0.5b^2 = 4 \implies b^2 = 8 = 4 * 2 \implies b = 2\sqrt{2} \implies$

$a = 2 * 8 = 16 \text { and } c = \sqrt[4]{0.5 * 8} = \sqrt{2}.$

And $16 + 8 + 4 = 28 \text { and } 16 * 2\sqrt{2} * \sqrt{2} = 16 * 2 * 2 = 64.$

$\therefore abc \le 64.$

7. ## Re: inequality

thank you sir

8. ## Re: inequality

Originally Posted by aray
thank you sir
You are welcome. Sorry for initial mistake. Idea came to rescue.

9. ## Re: inequality

Originally Posted by JeffM
$a > 0,\ b > 0\, c > 0, \text { and } a + b^2 + e^4 = 28.$

$z = abc - y(28 - a - b^2 - c^2) \implies$

$\color{red}{\dfrac{\delta z}{\delta a} = bc - y \implies bc = y \text { if } \dfrac{\delta z}{\delta a} = 0.}$

$\color{red}{\dfrac{\delta z}{\delta b} = ac - 2by \implies ac = 2by \text { if } \dfrac{\delta z}{\delta b} = 0.}$

$\color{red}{\dfrac{\delta z}{\delta c} = ab - 4c^3y \implies ab = 4c^3y \text { if } \dfrac{\delta z}{\delta c} = 0.}$

$\dfrac{\delta z}{\delta y} = 28 - a - b^2 - c^4 \implies a + b^2 + c^4 = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

$bc = y \implies y > 0 \text { and } c = \dfrac{y}{b}.$

$ac = 2by \implies a * \dfrac{y}{b} = 2by \implies a = 2b^2.$

$ab = 4c^3y \implies 2b^3 = 4c^3 * bc \implies 0.5b^2 = c^4 \implies c = \sqrt[4]{0.5b^2}.$

$\therefore 28 = a + b^2 + c^4 = 2b^2 + b^2 + 0.5b^2 = 3.5b^2 \implies$

$0.5 * 7b^2 = 28 \implies 0.5b^2 = 4 \implies b^2 = 8 = 4 * 2 \implies b = 2\sqrt{2} \implies$

$a = 2 * 8 = 16 \text { and } c = \sqrt[4]{0.5 * 8} = \sqrt{2}.$

And $16 + 8 + 4 = 28 \text { and } 16 * 2\sqrt{2} * \sqrt{2} = 16 * 2 * 2 = 64.$

$\therefore abc \le 64.$
For the three lines in red, it appears a negative sign was missed. It does not affect the end result (as the negatives seem to come in pairs, bringing the result back to positive).

$\dfrac{\partial z}{\partial a} = bc+y \Longrightarrow bc = -y \text{ if } \dfrac{\partial z}{\partial a} = 0$

$\dfrac{\partial z}{\partial b} = ac+2by \Longrightarrow ac = -2by \text{ if } \dfrac{\partial z}{\partial b} = 0$

$\dfrac{\partial z}{\partial a} = ab+4c^3y \Longrightarrow ab = -4c^3y \text{ if } \dfrac{\partial z}{\partial c} = 0$

$bc = -y \Longrightarrow y<0$ and $c=-\dfrac{y}{b}$.

$ac=-2by \Longrightarrow -a\dfrac{y}{b} = -2by \Longrightarrow a=2b^2$

$ab=-4c^3y \Longrightarrow 2b^3=-4c^3(-bc) \Longrightarrow 0.5b^2=c^4 \Longrightarrow c = \sqrt[4]{0.5b^2}$

The rest follows the same.

10. ## Re: inequality

Originally Posted by SlipEternal
For the three lines in red, it appears a negative sign was missed. It does not affect the end result (as the negatives seem to come in pairs, bringing the result back to positive).

$\dfrac{\partial z}{\partial a} = bc+y \Longrightarrow bc = -y \text{ if } \dfrac{\partial z}{\partial a} = 0$

$\dfrac{\partial z}{\partial b} = ac+2by \Longrightarrow ac = -2by \text{ if } \dfrac{\partial z}{\partial b} = 0$

$\dfrac{\partial z}{\partial a} = ab+4c^3y \Longrightarrow ab = -4c^3y \text{ if } \dfrac{\partial z}{\partial c} = 0$

$bc = -y \Longrightarrow y<0$ and $c=-\dfrac{y}{b}$.

$ac=-2by \Longrightarrow -a\dfrac{y}{b} = -2by \Longrightarrow a=2b^2$

$ab=-4c^3y \Longrightarrow 2b^3=-4c^3(-bc) \Longrightarrow 0.5b^2=c^4 \Longrightarrow c = \sqrt[4]{0.5b^2}$

The rest follows the same.
Umm

I set up z = abc MINUS y * (28 and so on).

As you say it makes no difference.

11. ## Re: inequality

Originally Posted by JeffM
Umm

I set up z = abc MINUS y * (28 and so on).

As you say it makes no difference.
Correct, $abc - y(28-a-b^2-c^4) = abc-28y \color{red}{+} ay \color{red}{+} b^2y \color{red}{+} c^4y$ because of the distributive property.

12. ## Re: inequality

We can also solve the problem using AM-GM

$\displaystyle \frac{\frac{a}{4}+\frac{a}{4}+\frac{a}{4}+\frac{a} {4}+\frac{b^2}{2}+\frac{b^2}{2}+c^4}{7}\geq \sqrt[7]{\frac{a^4 b^4 c^4}{4^42^2}}$