How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$

Question

Let $a,b,c,d>0$, show that

$$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$

I know this is interesting inequality,and Mathlinks can't solution,Thank you for you help

also can see this:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=595079

today,when I see this secrets in inequalitys Volume 1,(Pham kim Hung) page 205

let $a,b,c,d>0$ prove that $$\sum_{cyc}\dfrac{a^2}{b}\ge 2\sqrt{2}\sqrt[4]{a^4+b^4+c^4+d^4}$$

I can't prove it,can you help me? Thank you

Answer 1

Question:

Given $a,b,c,d > 0$, show that

$$ \frac{1}{4} \left( \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \right) \ge \sqrt[4]{ \frac{a^4 + b^4 + c^4 + d^4 }{4} } $$

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Write

$$ Q(x_1,x_2,x_3,x_4) = \frac{1}{4} \left( \frac{x_1^2}{x_2} + \frac{x_2^2}{x_3} + \frac{x_3^2}{x_4} + \frac{x_4^2}{x_1} \right) - \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^\tfrac{1}{4} $$

Find extreme value for $Q$:

$$ \frac{\partial Q}{\partial x_k} = \frac{1}{2} \frac{x_k}{x_{|k+1|}} - \frac{1}{4} \frac{x_{|k-1|}^2}{x_k^2} - \frac{1}{4} x_k^3 \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^{-\tfrac{3}{4}} = 0,\tag{1} $$

where $x_{|k\pm1|}$ is the cyclic index, thus $x_{|0|} = x_4$ and $x_{|5|} = x_1$. The extreme value is obtained for $x_1=x_2=x_3=x_4=x$ and

$$ Q(x,x,x,x) = 0. $$

We also have

$$ \frac{\partial^2 Q}{\partial x_k^2} = \frac{1}{2} \frac{1}{x_{|k+1|}} + \frac{1}{2} \frac{x_{|k-1|}^2}{x_k^3} + \frac{3}{16} x_k^6 \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^{-\tfrac{7}{4}} - \frac{3}{4} x_k^2 \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^{-\tfrac{3}{4}}, $$

and for $x_1=x_2=x_3=x_4=x$ we obtain

$$ \frac{\partial^2 Q}{\partial x_k^2} = \frac{7}{16} \frac{1}{x} \ge 0, $$

therefore

$$ Q(x_1,x_2,x_3,x_4) \ge 0, $$

whence

$$\frac{1}{4} \left( \frac{x_1^2}{x_2} + \frac{x_2^2}{x_3} + \frac{x_3^2}{x_4} + \frac{x_4^2}{x_1} \right) \ge \left( \frac{ x_1^4 + x_2^4 + x_3^4 + x_4^4}{4} \right)^\tfrac{1}{4}. $$

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To show that the stationary solution is unique.

From equation (1) follows that

$$ 2 \frac{x_{k+1}}{x_{k+2}} - \frac{x_{k}^2}{x_{k+1}^2} = \left( 2 \frac{x_{k+2}}{x_{k+3}} - \frac{x_{k+1}^2}{x_{k+2}^2} \right) \frac{x_{k+1}^3}{x_{k+2}^3}. $$

Let us define

$$ \xi_k = \frac{x_k}{x_{k+1}}, $$

then we obtain

$$ 2 \xi_{k+1} - \xi_k^2 = \Big( 2 \xi_{k+2} - \xi_{k+1}^2 \Big) \xi_{k+1}^3, $$

and it is clear that $\xi_k = 1$ is a solution. Let us write

$$ \xi_k = 1 + \phi_k, $$

then we obtain

$$ \phi_{k+2} = \frac{ \phi_{k+1}^5 + 5 \phi_{k+1}^4 + 10 \phi_{k+1}^3 + 8 \phi_{k+1}^2 + 3 \phi_{k+1} - \phi_{k}^2 - 2 \phi_k}{ \Big( 1 + \phi_{k+1} \Big)^2 }.\tag{2} $$

Note that for $\phi_\jmath > -1$, we obtain

$$ \begin{eqnarray} \textrm{sgn}\Big( \phi_{k+1}^5 + 5 \phi_{k+1}^4 + 10 \phi_{k+1}^3 + 8 \phi_{k+1}^2 + 3 \phi_{k+1} \Big) &=& \textrm{sgn}\Big(\phi_{k+1}\Big),\\ \textrm{sgn}\Big( \phi_{k}^2 - 2 \phi_k \Big) &=& \textrm{sgn}\Big(\phi_k\Big), \end{eqnarray} $$

whence we have the properties

$$ \begin{array}{ccccc} \phi_k \le 0 &\wedge& \phi_{k+1} \ge 0 &\Rightarrow& \phi_{k+2} \ge 0,\\ \phi_k \ge 0 &\wedge& \phi_{k+1} \le 0 &\Rightarrow& \phi_{k+2} \le 0.\\ \end{array} $$

We can use these properties to find "valid cycles":

$$ \begin{array}{ccccccccccc} \phi_k = 0 && \phi_{k+1} = 0 &\Rightarrow& \phi_{k+2} = 0 &\Rightarrow& \phi_{k+3} = 0\\ \phi_k = 0 && \color{red}{\phi_{k+1} < 0} &\Rightarrow& \phi_{k+2} < 0 && \phi_{k+3} < 0 && \phi_k = 0 &\Rightarrow& \color{red}{\phi_{k+1} > 0}\\ \color{red}{\phi_k = 0} && \phi_{k+1} < 0 &\Rightarrow& \phi_{k+2} < 0 && \phi_{k+3} > 0 &\Rightarrow& \color{red}{\phi_k > 0}\\ \color{red}{\phi_k = 0} && \phi_{k+1} > 0 &\Rightarrow& \phi_{k+2} > 0 && \phi_{k+3} < 0 &\Rightarrow& \color{red}{\phi_k < 0}\\ \phi_k = 0 && \color{red}{\phi_{k+1} > 0} &\Rightarrow& \phi_{k+2} > 0 && \phi_{k+3} > 0 && \phi_k = 0 &\Rightarrow& \color{red}{\phi_{k+1} < 0}\\ \hline \phi_k < 0 && \phi_{k+1} < 0 && \phi_{k+2} < 0 && \phi_{k+3} < 0\\ \color{red}{\phi_k < 0} && \phi_{k+1} < 0 && \phi_{k+2} < 0 && \phi_{k+3} > 0 &\Rightarrow& \color{red}{\phi_k > 0}\\ \phi_k < 0 && \phi_{k+1} < 0 && \phi_{k+2} > 0 &\Rightarrow& \phi_{k+3} > 0\\ \phi_k < 0 && \color{red}{\phi_{k+1} > 0} &\Rightarrow& \phi_{k+2} > 0 && \phi_{k+3} > 0 && \phi_{k} < 0 &\Rightarrow& \color{red}{\phi_{k+1} < 0}\\ \phi_k > 0 && \color{red}{\phi_{k+1} < 0} &\Rightarrow& \phi_{k+2} < 0 && \phi_{k+3} < 0 && \phi_{k} > 0 &\Rightarrow& \color{red}{\phi_{k+1} > 0}\\ \phi_k > 0 && \phi_{k+1} < 0 &\Rightarrow& \phi_{k+2} < 0 && \phi_{k+3} > 0 &\Rightarrow& \phi_k > 0\\ \phi_k > 0 && \phi_{k+1} > 0 && \phi_{k+2} < 0 &\Rightarrow& \phi_{k+3} < 0\\ \color{red}{\phi_k > 0} && \phi_{k+1} > 0 && \phi_{k+2} > 0 && \phi_{k+3} < 0 &\Rightarrow& \color{red}{\phi_k < 0}\\ \phi_k > 0 && \phi_{k+1} > 0 && \phi_{k+2} > 0 && \phi_{k+3} > 0\\ \end{array} $$

Therefore we only have the following valid cycles

$$ \begin{array}{cccc} 0&0&0&0\\ -&-&-&-\\ -&-&+&+\\ +&+&+&+\\ \end{array} $$

However the cycles

$$ \begin{array}{cccc} -&-&-&-\\ +&+&+&+\\ \end{array} $$

can be excluded for $\xi_1 \xi_2 \xi_3 \xi_4 = 1$.

What is left to show is that the cycle $--++$ leads to contradiction.

To be continued...

Answer 2

The above answer by johannesvalks "is not yet completed", as johannesvalks himself points out. Here is a new answer without calculus.

Rewrite the question as follows: Show that $$ f = (a b c d)^4\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4 - 64 (a b c d)^4\left({{a^4+b^4+c^4+d^4}}\right) \ge 0 $$

Due to cyclicity, we can demand that $d = \min\{a,b,c,d\}$. Due to homogeneity, we can demand that $d=1$. Hence we need to show $$ f = \left(a^3c+ab^3 + abc^3 + bc\right)^4 - 64 (a b c)^4\left({{a^4+b^4+c^4+1}}\right) \ge 0 $$ Now this has become an inhomogeneous polynomial of degree 20.

We now distinguish cases for $a,b,c$. As $1 = d = \min\{a,b,c,d\}$, we write $a = 1+A$, $b = 1+B$, $c = 1+C$, with $A,B,C \ge 0$. Indeed, we are not going to use $A,B,C$. When doing so, and fully expanding (by computer) $f$, we obtain terms with negative sign, which (without further processing) makes the result inconclusive.

Remark: "Standard" Buffalo Way doesn't succeed here, since we have to construct all 6 cases of the ordering of $A,B,C$. In the two cases where we let $a=1+x$; $c = 1+x+y$; $b=1+x+y+z$, and $c=1+x$; $a = 1+x+y$; $b=1+x+y+z$, with $x,y,z \ge 0$, we obtain after fully expanding (by computer) $f$, terms with negative sign.

Instead, we consider the following 4 cases:

Case 1: $A<B+C$; $B<A+C$; $C<A+B$. Visually, these three conditions are met if $A,B,C$ are "similar" in value. In this case, we can write $A = x+y$, $B= y+z$, $C= x+z$, with $x,y,z \ge 0$. This is possible, since $x = \frac{A+C-B}{2}$, and cyclically for $y,z$. Inserting $a = 1 + x+y$ etc. into $f$ and fully expanding (by computer) gives the expression $f_1$ in the appendix below, which clearly is nonnegative since all terms have positive sign.

Further cases: Note that for $A,B,C \ge 0$, only one of the conditions $A<B+C$; $B<A+C$; $C<A+B$ can be violated. Proof: suppose the first two conditions indeed are reverted to $A \ge B +C$ and $B\ge A+C$; then we add these two conditions and have $C \le 0$ which is a contradiction. The same argument holds cyclically. Hence we only need to consider one reverted inequality in $A,B,C$ (3 more cases). This covers all possible cases.

Case 2: Visually, this condition is met if $a$ is "largest" in value. $A \ge B +C$; $B<A+C$; $C<A+B$. Writing $B=x$; $C = y$; $A = x+y+z$, with $x,y,z \ge 0$, satisfies these conditions. Inserting $a = 1 + x+y+z$, $b = 1 + x$, $c = 1 + y$ into $f$ and fully expanding (by computer) gives the expression $f_2$ in the appendix below, which clearly is nonnegative since all terms have positive sign.

Case 3: As case 2, with $b$ the largest value. Inserting $b = 1 + x+y+z$ etc. into $f$ and fully expanding (by computer) gives the expression $f_3$ in the appendix below, which clearly is nonnegative since all terms have positive sign.

Case 4: As case 2, with $c$ the largest value. Inserting $c = 1 + x+y+z$ etc. into $f$ and fully expanding (by computer) gives the expression $f_4$ in the appendix below, which clearly is nonnegative since all terms have positive sign.

This proves the claim. $\qquad \Box$

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Appendix: The four cases are given as a diary file from MATLAB (The "*"s have been removed). The diary comes as files, as the number of letters exceeds the limit allowed to be put in directly as text. MATLAB diary allows to re-perform the calculations, if wished. Obviously, the results are VERY lengthy, however the relevant fact to notice is that there are no terms with negative signs in the expanded versions of $f_1, f_2, f_3, f_4$, which can safely be verified by actually searching the files for minus signs (and not finding any in the expanded versions).