Minkowski Inequality

I will be trying to post a Minkowski inequality solution, till then lets look at this.

$f(a,b,c) = \sqrt[3]{a^3+\dfrac{1}{b^3}}+\sqrt[3]{b^3+\dfrac{1}{c^3}}+\sqrt[3]{c^3+\dfrac{1}{a^3}} \text{ subject to } a+b+c\le\frac{3}{2}$

Using partial differentiation i.e solving $\frac{\partial{f(a,b,c)}}{\partial{a}} , \frac{\partial{f(a,b,c)}}{\partial{b}},\frac{\partial{f(a,b,c)}}{\partial{c}} = 0$

We have the equations :

$\begin{aligned} a^3+(a^2c)^3 - (ab)^{-3} = b^3+(b^2a)^3 - (bc)^{-3} = c^3+(c^2b)^3 - (ca)^{-3} = \frac{a^3+(a^2c)^3 - (ab)^{-3} + b^3+(b^2a)^3 - (bc)^{-3} + c^3+(c^2b)^3 - (ca)^{-3}}{3}\end{aligned}$

Since the combined equation is homogeneous in $(a,b,c)$ by WLOG we have $a\le b\le c$ & for minimum we must have $a\le b \le c \le \frac{a+b+c}{3} \le \frac{1}{2}$

Putting the minimum value of $a,b,c$ in the expression we get maximum as $\boxed{6.031}$

Relevant wiki: Lagrange Multipliers

It is easy to see with the Minkowski inequality that: $f(a,b,c)=\sqrt [ 3 ]{ { a }^{ 3 }+\frac { 1 }{ { b }^{ 3 } } } +\sqrt [ 3 ]{ b^{ 3 }+\frac { 1 }{ c^{ 3 } } } +\sqrt [ 3 ]{ c^{ 3 }+\frac { 1 }{ a^{ 3 } } } \ge \sqrt [ 3 ]{ { (a+b+c) }^{ 3 }+{ (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }^{ 3 } } =g(a,b,c)$

Taking the directional derivative of $g(a,b,c)$ allong $\overrightarrow { v } =\begin{bmatrix} { 3 }^{ -1/2 } \\ { 3 }^{ -1/2 } \\ { 3 }^{ -1/2 } \end{bmatrix}$ , we get: $\sum _{ i=1 }^{ 3 }{ \frac { 1 }{ \sqrt { 3 } } \frac { 1 }{ 3{(g(a,b,c))}^{2} } \left[ 3{ (a+b+c) }^{ 2 }-\frac { 3 }{ { x }_{ i }^{ 2 } } { (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }^{ 2 } \right] }$

Where ${x}_{1}=a, {x}_{2}=b, {x}_{3}=c$ . We want to show that this is negative for $a+b+c \le 3/2$ . Exploiting the fact that $g(a,b,c)>0$ , for our condition to hold we get: $\sum _{ i=1 }^{ 3 }{ { x }_{ i } } <\sum _{ i=1 }^{ 3 }{ (\frac { 1 }{ { x }_{ i } } \sum _{ j=1 }^{ 3 }{ \frac { 1 }{ { x }_{ j } } ) } } ={ \left( \sum _{ j=1 }^{ 3 }{ \frac { 1 }{ { x }_{ j } } } \right) }^{ 2 }$

Assuming $\sum _{ i=1 }^{ 3 }{ { x }_{ i } } =y$ , we get that the maximum value of the smallest ${x}_{i}$ is $y/3$ . This implies that ${\left( \sum_{ j=1 }^{ 3 }{ \frac { 1 }{ { x }_{ j } } } \right) }^{ 2 } > \frac { 1 }{ { x }_{ n }^{ 2 } } \ge \frac { 9 }{ { y }^{ 2 } }$ . Therefore our condition definitely holds for $y<\frac { 9 }{ { y }^{ 2 } } \Rightarrow y<{9}^{1/3}\approx 2.08$ .

From this we can conclude that the minimum of $g(a,b,c)$ must lie on $a+b+c=3/2$ We now create the lagrange multiplier of ${g(a,b,c)}^{3}$ , which has the same minimum as $g(a,b,c)$ , with the constraint $a+b+c=3/2$ . $L (a,b,c,\lambda )={ (a+b+c) }^{ 3 }-{ (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }^{ 3 }-\lambda (a+b+c-3/2)=27/8-{ (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }^{ 3 }-\lambda (a+b+c-3/2)$ Taking the derivative with respect to ${x}_{i}$ : ${ \partial }_{ { x }_{ i } }(L({ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },\lambda))=3{ (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }^{ 2 }\frac { 1 }{ { x }_{ i }^{ 2 } } -\lambda$

Since ${ \partial }_{ { x }_{ i } }(L({ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },\lambda))=0$ at the minimum: $3{ (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }^{ 2 }\frac { 1 }{ { x }_{ i }^{ 2 } } -\lambda =0$ ${ (\frac { 1 }{ { a } } +\frac { 1 }{ { b } } +\frac { 1 }{ c } ) }\sqrt { 3/\lambda } ={ x }_{ i }$ $a=b=c=1/2$ Therefore the minimum value of $g(a,b,c)$ is $\sqrt [ 3 ]{ 1755 } /2$ . If we evaluate $f(1/2,1/2,1/2)$ , we get the same value.

Since $g(a,b,c)\le f(a,b,c)$ , the minimum value of $f(a,b,c)$ is $\sqrt [ 3 ]{ 1755 } /2\approx 6.031$ .

Let

$x = a + b + c, y = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ .

Using Minkowski inequality we get:

$\sqrt[3]{ a^3 + \frac{1}{b^3} } + \sqrt[3]{ b^3 + \frac{1}{c^3} } + \sqrt[3]{ c^3 + \frac{1}{a^3} } \ge \sqrt[3]{ (a + b + c)^3 +\big( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} \big)^3 } = \sqrt[3]{ x^3 +y^3 }$

From the statement $x \le 3/2$ .

From Arithmetic-Harmonic Mean Inequality we get:

$\frac{3}{y} \le \frac{x}{3} \le {1/2}$

From here we can find out that $xy \ge 9$ and $y \ge 6$

Now let us use Cauchy inequality for 65 terms:

$x^3 + 64 \cdot \frac{1}{64} y^3 \ge 65 \sqrt[65] { x^3 \cdot \big( \frac{y}{4} \big)^{64 \cdot 3} } = 65 \sqrt[65] { \big(\frac{xy}{4}\big)^3 \cdot \big( \frac{y}{4} \big)^{63 \cdot 3} } \ge 65 \sqrt[65] { \big(\frac{9}{4}\big)^3 \big( \frac{6}{4} \big)^{63 \cdot 3} } = 65 \cdot \big( \frac{3}{2} \big)^3$

(It may be unclear why do I take exactly these terms. We want equality to be achieved in Cauchy inequality so all terms should be equal and since intuitively $a = b = c = 1/2$ in minimum, then we should use inequality for the equal terms: $x^3, (y/4)^3, (y/4)^3, ...$ . This is of course informal explanation.)

Now we finally get

$\sqrt[3]{ a^3 + \frac{1}{b^3} } + \sqrt[3]{ b^3 + \frac{1}{c^3} } + \sqrt[3]{ c^3 + \frac{1}{a^3} } \ge \sqrt[3] { x^3 + y^3 } \ge \frac{3}{2} \sqrt[3] { 65} \approx \boxed{6.031}$

From the Minkowski inequality we get that equality holds if and only if $a = b = c = \frac{1}{2}$ .

Set the polynomial is P By applying C-S consequence(Mincowski) inequality: $P\geq \sqrt[3]{(x+y+z)^{3}+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^3}\geq \sqrt[3]{27xyz+\frac{1}{27xyz}}$ Set $xyz=t;0< t\leq \frac{1}{2}$ We see $f(t)$ is decreasing in that interval. So $min f(t)=f(\frac{1}{8})=\frac{1755}{8}$

So $P\geq \sqrt[3]{\frac{1755}{8}}\approx 6.031$ .

By Minkowski's Inequality for $p<1$ , we have

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

By the Cauchy–Schwarz Inequality or Arithmetic-Harmonic Mean Inequality, it is easy to show that:

$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3 \geq 216$

$\left(\sqrt[3]{a^3+\frac{1}{b^3}}+\sqrt[3]{b^3+\frac{1}{c^3}}+\sqrt[3]{c^3+\frac{1}{a^3}}\right)^3 \geq (a+b+c)^3 + 216$

Note that this occurs only when $a=b=c=\frac{1}{2}$

Also note that $a+b+c$ is maximised at the same values

$\left(\sqrt[3]{a^3+\frac{1}{b^3}}+\sqrt[3]{b^3+\frac{1}{c^3}}+\sqrt[3]{c^3+\frac{1}{a^3}}\right)^3 \geq \frac{27}{8} + 216$

Unfortunately, the above logic is not correct, and therefore, invalid, even though it is the true minimum of the expression. I presently do not see a viable means of justifying the above conclusion.

Discuss in the comments.

Since a lot of people have complained about this asking for minimum in stead of maximum, I went to check very carefully and realized my mistake. Therefore, a (not at all) slight modification has been made. Sorry for all the inconvenience!