Inequality with wx + xy + yz + zw = 1

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Inequality with wx + xy + yz + zw = 1

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4-variable inequality

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Source: IMO ShortList 1990, Problem 24 (THA 2)

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444 posts

#1 h Nov 2, 2005, 2:38 PM • 4 Y

Y by uvwmethod, Adventure10, megarnie, and 1 other user

Let $w, x, y, z$ are non-negative reals such that $wx + xy + yz + zw = 1$ .
Show that $\frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$ .

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#2 h Nov 2, 2005, 2:52 PM • 5 Y

Y by Adventure10, Adventure10, Mango247, ehuseyinyigit, and 1 other user

More generaly we prove that if $a_1,\cdots,a_n$ are positive real numbers with sum $s$ and such that $a_1a_2+\cdots+a_na_1=k^2$ , then for $r=0,1,2,3$ we have:
$s_n= {{a_1^r}\over{s-a_1}}+{{a_2^r}\over{s-a_2}}+\cdots+{{a_n^r}\over{s-a_n}}\geq {{(2k)^{r-1}}\over{ (n-1)n^{r-2}}}.$
Indeed, the function $x\mapsto {{x^r}\over{s-x}}$ is convex for $s>x\geq 0$ and $r=0,1,2,3$ . Hence by Jensen' inequality we have
$s_n\geq {{n(s/n)^r}\over{s-s/n}}={{s^{r-1}}\over{(n-1)n^{r-2}}}.$
On the other hand, we know that $(a_1+\cdots+a_n)^2\geq 4(a_1a_2+\cdots+a_na_1)=4k^2$ and thus
$s_n\geq {{(2k)^{r-1}}\over{(n-1)n^{r-2}}}.$
The stated problem corresponds to: $n=4, r=3$ and $k=1$ .

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altusi

#3 h Nov 2, 2005, 4:01 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

or to write: x^3/(y+z+w)=x^4/x(z+y+w) for all, then use cauchy-schwarz! we will arrive to a simple thing...

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arqady

#4 h Nov 4, 2005, 9:40 AM • 7 Y

Y by N0_NAME, Adventure10, Mango247, kiyoras_2001, and 3 other users

$(w-x)^2+(x-y)^2+(y-z)^2+(z-w)^2\geq0.$
Hence, $w^2+x^2+y^2+z^2\geq1.$
$\frac{a^3}{b}\geq2a^2-ab,$ where $a>0,b>0.$
Hence, $\sum\frac{(3w)^3}{x+y+z}\geq3(6\sum w^2-2\sum wx)\geq3(5-2+(w-y)^2+(x-z)^2)\geq9.$
Hence, $\sum\frac{w^3}{x+y+z}\geq\frac{1}{3}.$

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#5 h Sep 25, 2007, 3:35 AM • 1 Y

Y by Adventure10

Problem:Let $a;b;c;d$ be positive real numbers satisfying $ab+bc+ca+da=1$ .Prove that:
$\sum\frac{a^{3}}{b+c+d}\geq\frac{1}{3}$
My solution
By AM-GM we have:
$\frac{a^{3}}{b+c+d}{18}+\frac{a}{6}+\frac{1}{12}\geq\frac{2a}{3}$
Similar,we have $\sum\frac{a^{3}}{b+c+d}\geq\frac{a+b+c+d}{3}-\frac{1}{3}$
Note that $ab+bc+ca+da=(a+c)(b+d)$ ,so $(a+b+c+d)^{2}\geq 4(ab+bc+cd+da)=4$ => $a+b+c+d\geq 2$
Now,we have $\sum\frac{a^{3}}{b+c+d}\geq\frac{1}{3}$
Done!

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#6 h Sep 25, 2007, 5:06 AM • 2 Y

Y by Adventure10, Mango247

Other problem:
Let $a;b;c;d$ be positive real numbers.Prove that:
$\frac{a^{3}+b^{3}+c^{3}}{b+c+d}\geq\sum a^{2}$

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orl

#7 h Aug 15, 2008, 4:31 PM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Let let $a$ , $b$ , $c$ and $d$ positive reals such us $ab + bc + cd + da = 1$ .
Show that ${\frac {{a}^{3}}{b + c + d}} + {\frac {{b}^{3}}{c + d + a}} + {\frac {{c}^{3}} {a + b + c}} + {\frac {{d}^{3}} {d + c + a}}\ge1/3$

Approach by campos:

by hölder we have that

$\left(\sum \frac {a^3}{b + c + d}\right) \left(\sum a(b + c + d)\right) \left(\sum 1\right)^2\geq \left(\sum a\right)^4 \\ \\ \Rightarrow \sum \frac {a^3}{b + c + d}\geq \frac {(a + b + c + d)^4}{16(\sum a(b + c + d))}$

but it's easy to verify from the condition that
$(a + b + c + d)^2\geq 4(a + c)(b + d) = 4$
and
$3(a + b + c + d)^2\geq 4(2 + 2ac + 2bd) = 4((a + b + c + d)^2 - (a^2 + b^2 + c^2 + d^2)) \\ \Leftrightarrow 4(a^2 + b^2 + c^2 + d^2)\geq (a + b + c + d)^2$
which is true by cauchy-schwarz

Approach by Sung-yoon Kim:

cronecker wrote:

Let $a$ , $b$ , $c$ and $d$ positive reals such us $ab + bc + cd + da = 1$ .
Show that ${\frac {{a}^{3}}{b + c + d}} + {\frac {{b}^{3}}{c + d + a}} + {\frac {{c}^{3}} {a + b + c}} + {\frac {{d}^{3}} {d + c + a}}\ge1/3$

I think the original form is like this : $\sum {\frac {a^3}{b + c + d}} \geq \frac {1}{3}$ . By Cauchy, $(LHS)(a(b + c + d) + b(c + d + a) + c(d + a + b) + d(a + b + c)) \geq (a^2 + b^2 + c^2 + d^2)^2$ . So $(LHS)\geq {\frac {(a^2 + b^2 + c^2 + d^2)^2}{2(ab + ac + ad + bc + bd + cd)}}$ . -(*)

By AM-GM, $a^2 + b^2 + c^2 + d^2 \geq 2(ab + cd), 2(ac + bd), 2(ad + bc)$ . So $(a^2 + b^2 + c^2 + d^2)\geq ab + bc + cd + da$ and $3(a^2 + b^2 + c^2 + d^2) \geq ab + ac + ad + bc + bd + cd$ . Multiplying this, we get that the RHS of (*) is not less than 1/3.

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#8 h Aug 15, 2008, 8:52 PM • 2 Y

Y by Adventure10, Mango247

Fermat -Euler wrote:

Let $w, x, y, z$ are non-negative reals such that $wx + xy + yz + zw = 1$ .
Show that $\frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$ .

Follows Holder Inequality we have

$LHS\ge\frac{(x+y+z+w)^3}{4\cdot3(x+y+z+w)}=\frac{(x+y+z+w)^2}{4\cdot3(xy+yz+zw+wx)}\ge\frac{1}{3}$

because $(x-y+z-w)^2\ge 0$

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manlio

#9 h Aug 27, 2008, 10:01 AM • 2 Y

Y by Adventure10, Mango247

chien than wrote:

Other problem:
Let $a;b;c;d$ be positive real numbers.Prove that:
$\frac {a^{3} + b^{3} + c^{3}}{b + c + d}\geq\sum a^{2}$

Sorry I cannot solve it and I don't remember if it was already posted.

chien than wrote:

Problem:Let $a;b;c;d$ be positive real numbers satisfying $ab + bc + ca + da = 1$ .Prove that:
$\sum\frac {a^{3}}{b + c + d}\geq\frac {1}{3}$
My solution
By AM-GM we have:
$\frac {a^{3}}{b + c + d}{18} + \frac {a}{6} + \frac {1}{12}\geq\frac {2a}{3}$

How to correct AM-GM ?

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#10 h Aug 27, 2008, 12:36 PM • 2 Y

Y by Adventure10, Mango247

manlio wrote:

chien than wrote:

Other problem:
Let $a;b;c;d$ be positive real numbers.Prove that:
$\frac {a^{3} + b^{3} + c^{3}}{b + c + d}\geq\sum a^{2}$

Sorry I cannot solve it and I don't remember if it was already posted.

chien than wrote:

Problem:Let $a;b;c;d$ be positive real numbers satisfying $ab + bc + ca + da = 1$ .Prove that:
$\sum\frac {a^{3}}{b + c + d}\geq\frac {1}{3}$
My solution
By AM-GM we have:
$\frac {a^{3}}{b + c + d}{18} + \frac {a}{6} + \frac {1}{12}\geq\frac {2a}{3}$

How to correct AM-GM ?

Dear manlio,
It should be
$\sum \frac{a^3+b^3+c^3}{b+c+d} \ge \sum a^2$
This inequality is easy with AM-GM Inequality, since
$\frac{a^3}{b+c+d} +\frac{a(b+c+d)}{9} \ge \frac{2a^2}{3}$
$\Rightarrow \sum \frac{a^3}{b+c+d} \ge \frac{2}{3}\sum a^2 -\frac{2}{9}(ab+ac+ad+bc+bd+cd) \ge \frac{2}{3}\sum a^2-\frac{1}{3} \sum a^2 =\frac{1}{3}\sum a^2$
Similarly, also by AM-GM,
$\frac{b^3}{b+c+d} +\frac{b(b+c+d)}{9} \ge \frac{2}{3}b^2$
$\Rightarrow \sum \frac{b^3}{b+c+d} \ge \frac{2}{3}\sum a^2 -\frac{1}{9}\sum a^2 -\frac{1}{9} (a+c)(b+d) -\frac{2}{9}(ac+bd) \ge \frac{2}{3}\sum a^2 -\frac{1}{9}\sum a^2 -\frac{1}{9}\sum a^2 -\frac{1}{9}\sum a^2 =\frac{1}{3}\sum a^2$
Similarly, we also have
$\sum \frac{c^3}{b+c+d} \ge \frac{1}{3}\sum a^2$
Adding up these inequalities, we have done.

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#11 h Aug 28, 2008, 12:46 AM • 2 Y

Y by Adventure10, Mango247

Exist a proof which is used Chebysev ineq ,Can_hang 2007.but your solution is very nice

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manlio

#12 h Aug 28, 2008, 9:22 AM • 2 Y

Y by Adventure10, Mango247

Thank you very can_hang2007 for your nice solution.

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#13 h Jun 22, 2013, 3:41 PM • 3 Y

Y by megarnie, Adventure10, Mango247

The given condition implies $1 = wx+xy+yz+zx \le w^2+x^2+y^2+z^2$ . Then by Cauchy, $\sum_{\text{cyc}} \frac{w^4}{wx+wy+wz} \ge \frac{(w^2+x^2+y^2+z^2)}{2(wx+xy+yz+zx+wy+xz)} \ge \frac{w^2+x^2+y^2+z^2}{2(wx+xy+yz+zx+wy+xz)} \ge \frac{1}{3}.$

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sayantanchakraborty

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sayantanchakraborty

#14 h May 19, 2014, 6:23 AM • 1 Y

Y by Adventure10

$\sum{\frac{x^3}{y+z+w}}=\sum{\frac{x^4}{xy+xz+xw}} \ge \frac{(\sum{x^2})^2}{2\sum{xy}} (\text{Titu's Lemma}) \ge \frac{2\sum{x^2}-1}{2\sum{xy}} ((\sum{x^2}-1)^2 \ge 0 \Leftrightarrow (\sum{x^2})^2 \ge 2\sum{x^2}-1) \ge \frac{2\sum{x^2}-1}{3\sum{x^2}}$

Now $\frac{2\sum{x^2}-1}{3\sum{x^2}} \ge \frac{1}{3}$
$\Leftrightarrow \sum{x^2} \ge 1$

But this follows with power-mean inequality combined with AM-GM:

$\frac{\sum{x^2}}{4} \ge (\frac{\sum{x}}{4})^2 \ge \frac{4(y+w)(x+z)}{16}=\frac{1}{4}$

from which the result follows...It is easy to see that equality holds if and only if $x=y=z=w=\frac{1}{2}$

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#15 h Aug 18, 2014, 7:47 PM • 2 Y

Y by Adventure10, Mango247

By cyclic AM-GM's $x^2+y^2+z^2+w^2\ge 1$ , and also $\sum_{\text{sym}} x^2\ge \tfrac{1}{3}\left( \sum _{\text{sym}} xy\right)$ . Hence by Cauchy
$\tfrac{w^4}{1-(x+z-w)y}+\tfrac{x^4}{1-(y+w-x)z}+\tfrac{y^4}{1-(x+z-y)w}+\tfrac{z^4}{1-(y+w-z)x}\ge$ $\ge\tfrac{(w^2+x^2+y^2+z^2)^2}{2+2yw+2xz}=\tfrac{\left( \sum_{\text{sym}} x^2\right) ^2}{\sum_{\text{sym}} xy}\ge \tfrac{\sum_{\text{sym}} x^2}{3}\ge \tfrac{1}{3}.$

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#16 h Dec 23, 2018, 7:36 PM • 2 Y

Y by Adventure10, Mango247

We use Titu's Lemma, or Fractional Cauchy-Schwarz. Forming fourth powers in the numerator and applying the Lemma,

$\sum_{\text{cyc}} \frac{w^4}{w(x+y+z)} \geq \frac{(w^2+x^2+y^2 + z^2)^2}{2(wx+wy+wz+xy+yz+zx)} \geq \frac{1}{3}$
With the given $wx+xy+yz + zw = 1$ , the denominator can be rearranged into $2(1+wy+xz)$ . Therefore, it remains to show that

$\frac{(w^2 + x^2 + y^2 + z^2)^2}{2(1+wy+xz)} \geq \frac{1}{3}$
It is sufficient to show that $3(w^2 + x^2 + y^2 + z^2) \geq 2(1+wy+xz)$ . By AM-GM, we have $w^2 + y^2 \geq 2wy$ and $x^2 + z^2 \geq 2xz$ , which yield $w^2 + y^2 + x^2 + z^2 \geq 2(xz + wy)$ when added together. Furthermore, by the well known fact that $a_1^2 + a_2^2 + a_3^2 + a_4^2 \geq a_1a_2 + a_2a_3 + a_3a_4 + a_4a_1$ , we have that $w^2 + x^2 + y^2 + z^2 \geq wx + xy + yz + zx =1$ . The conclusion follows.

(P.S: Sorry if a similar proof was already posted)

This post has been edited 2 times. Last edited by OmicronGamma, Dec 23, 2018, 7:50 PM

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InCtrl

#17 h Dec 5, 2019, 4:46 AM • 1 Y

Y by Adventure10

I think this is different enough to warrant a post. Oh well.

By T2 Lemma and the given we get to $\text{Expression}\ge \frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}$ .
If $wy+zx\le \frac{1}{2}$ , then we have, by the rearrangement ineq $w^2+x^2+y^2+z^2\ge wx+xy+zy+zx =1$ , $\frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}\ge \frac{1}{2+2(wy+xz)}\ge \frac13$ .

Otherwise, we have $u=wy+xz\ge \frac12$ .
By am-gm, $w^2+x^2+y^2+z^2\ge 2wy+2zx=2u$ , so our inequality rewrites as
$\frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}\ge \frac{(4u^2)}{2(1+u)}\ge\frac13\iff 6u^2-u-1\ge0\iff (3u+1)(2u-1)\ge 0$ which is true.

Combining both cases, we're done.

This post has been edited 1 time. Last edited by InCtrl, Dec 5, 2019, 4:47 AM

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Pumpko

#18 h Mar 14, 2020, 2:44 PM

Y by

Solved using the AM-GM inequality, English is not my native language
$(w+x+y+z)^2 \geq 4(w+y)(x+z) = 4(wx+xy+yz+zw) = 4$
$\Rightarrow w+x+y+z \geq 2$
We want to prove $\sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{1}{3}$ By AM-GM we have $\frac{w^3}{x+y+z}+\frac{x+y+z}{a}+\frac{1}{b} \geq \frac{3}{\sqrt[3]{ab}}w$
$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{3(w+x+y+z)}{\sqrt[3]{ab}}-\frac{3(w+x+y+z)}{a}-\frac{4}{b}$ $\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq 3(w+x+y+z)(\frac{1}{\sqrt[3]{ab}}-\frac{1}{a})-\frac{4}{b}$ We want $\frac{1}{\sqrt[3]{ab}}-\frac{1}{a}>0 \Rightarrow a^2>b$
$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{6}{\sqrt[3]{ab}}-\frac{6}{a}-\frac{4}{b}=\frac{1}{3}$ Let $\frac{1}{3}=\omega-\alpha-\beta$
$\Rightarrow \omega=\frac{6}{\sqrt[3]{ab}}, \alpha=\frac{6}{a}, \beta=\frac{4}{b}$
$\Rightarrow ab=\frac{216}{\omega^{3}}, a=\frac{6}{\alpha}, b=\frac{4}{\beta}$
$\Rightarrow 9\alpha\beta=\omega^{3}$
So any number $\omega, \alpha$ and $\beta$ that satisfy $\begin{cases} \omega-\alpha-\beta=\frac{1}{3}\\ 9\alpha\beta=\omega^{3}\\ a^2>b\end{cases}$ will give us the numbers $x, y$ to prove the inequality
For example
$\begin{cases}\omega=1\\\alpha=\frac{1}{3}\\\beta=\frac{1}{3}\\\end{cases}$ wil give us $\begin{cases}a=18\\b=12\end{cases}$ and we can use that to prove the original inequality
If we plug in any $a$ and $b$ that don't follow this "rule" we will get $\sum_{cycle} \frac{w^3}{x+y+z} \geq \gamma, \gamma < \frac{1}{3}$ This took me really long to figure this out, but when I did I was really happy

This post has been edited 1 time. Last edited by Pumpko, Mar 14, 2020, 10:34 PM
Reason: realized I forgot 1 condition

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sqing

#19 h Jun 27, 2020, 12:40 PM

Y by

Fermat -Euler wrote:

Let $w, x, y, z$ are non-negative reals such that $wx + xy + yz + zw = 1$ .
Show that $\frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$ .

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#20 h Dec 20, 2021, 3:30 AM

Y by

It is easy to use Titu's Lemma.
The equation, assume S
$S\geq \frac{(w^2+x^2+y^2+z^2)^2}{2(xw+yw+zw+xy+yz+xz)}$
Since $wx+yx+yz+zw=1, w^2+y^2+z^2+x^2\geq 1; 2xz+2yw\geq 1$
$(w^2+x^2+y^2+z^2)^2\geq (wx+xy+yz+zw)^2=1$ by C-S inequality
so $S\geq \frac{1}{2*1+1}=\frac{1}{3}$

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sqing

#21 h Jun 5, 2023, 9:24 AM

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Let $a,b,c,d$ are non-negative reals such that $a b+a c+a d+b c+bd+c d=1$ . Show that $\frac{a^3}{b+c+d}+\frac{b^3}{c+d+a}+\frac{c^3}{a+b+d}+\frac{d^3}{a+b+c}\geq \frac {2}{9}$

This post has been edited 1 time. Last edited by sqing, Jun 5, 2023, 9:24 AM

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Clew28

#22 h Jun 21, 2024, 2:52 AM • 2 Y

Y by GeoKing, duckman234

By the Cauchy-Schwarz inequality, we have: $\quad \sum_{\text{cyc}} \frac{a^3}{b+c+d} = \sum_{\text{cyc}} \frac{a^4}{ab+ac+ad} \geq \frac{a^2 + b^2 + c^2 + d^2}{\sum_{\text{cyc}} (ab + ac + ad)} \geq \frac{1}{3}$ and we are done

This post has been edited 1 time. Last edited by Clew28, Jun 23, 2024, 3:24 AM

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#23 h Jan 11, 2025, 1:49 AM

Y by

Use $a$ , $b$ , $c$ , and $d$ instead of $w$ , $x$ , $y$ , and $z$ , because I copied this from OTIS.

Let $s = a+b+c+d$ .

Claim: $\sum \frac{a^3}{s-a}\ge \frac{s^2}{12}.$
Since $f(x)=\frac{x^3}{s-x}$ is convex, Jensen's inequality gives $\frac{f(a)+f(b)+f(c)+f(d)}{4}\ge f(\frac{s}{4}).$ This is equivalent to $f(a)+f(b)+f(c)+f(d)\ge \frac{s^2}{12}\ge \frac{1}{3}.$ The last part is simple; if we define $x = a+c$ and $y = b+d$ , then $xy = 1$ so $x+y=s\ge 2.$

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#24 h Mar 16, 2025, 10:01 PM

Y by

The trick is to homogenize and dehomogenize the expression in such a way that the restriction on $w, x, y, z$ is nicer.

In particular, the homogenized version is $\begin{align*} \sum_{\text{cyc}} \frac{w^3}{x + y + z} \geq \frac{wx + xy + yz + zw}{3}. \end{align*}$
We dehomogenize and set $w + x + y + z = 4$ , and obtain $\begin{align*} \sum_{\text{cyc}} \frac{w^3}{4 - w} \geq \frac{wx + xy + yz + zw}{3}. \end{align*}$
This is actually a very weak inequality, and we can prove this easily using the tangent-line trick.

We first claim that for any nonnegative real $n$ , we have $\begin{align*} \frac{n^3}{4 - n} \geq \frac{10n}{9} - \frac{7}{9}. \end{align*}$
This rearranges to $\begin{align*} (n-1)^2(9n + 28) \geq 0, \end{align*}$
which is clearly true.

Now, we can finish the problem. Observe that $\begin{align*} \sum_{\text{cyc}} \frac{w^3}{4 - w} \geq \sum_{\text{cyc}} \frac{10w}{9} - \frac{7}{9} = \frac{4}{3}. \end{align*}$
Additionally, using AM-GM, we have $\begin{align*} wx + xy + yz + zw = (w+y)(x + z) \leq \left(\frac{(w + y)(x + z)}{2}\right)^2 = 4. \end{align*}$
Finally, $\begin{align*} \sum_{\text{cyc}} \frac{w^3}{4 - w} \geq \frac{4}{3} \geq \frac{wx + xy + yz + zw}{3}, \end{align*}$
and we are done.

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