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Question:

Let $$a,b,c,d>0,a+b+c+d=4$$

show that $$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$

when I solved this problem, I have see following three variables inequality:

Assumming that $a,b,c>0,a+b+c=3$, show that :$$f(a,b,c)=\dfrac{a}{a+bc}+\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\ge\dfrac{3}{2}$$

solution can see:inequality

I found this three answer all is not true,

1、such as dear @Macavity, in fact $$\sum_{cyc}\dfrac{a^2}{a^2+abc}\ge\dfrac{16}{\sum_{cyc}(a^2+abc)}$$ the Right not $\dfrac{16}{4+\sum_{cyc}abc}$

2、and the @ante.ceperic is also not true.in fact $$a^2+b^2+c^2+d^2+abc+bcd+cda+dab\le 8$$ is not true with $a+b+c+d=4$

such let $a=3$

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  • $\begingroup$ Missed your message, as the system will not notify if you edit it in like that. I see the error and will get back if I can correct it. $\endgroup$ – Macavity Dec 2 '14 at 7:31
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Let $$a,b,c,d>0,a+b+c+d=4$$

show that $$P=\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$

Now using Cauchy Schwarz: $$\left(\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\right)((a+bc)+(b+cd)+(c+da)+(d+ab))\ge (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$$ Now: $$P\ge\frac{4+2(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd})}{4+ab+bc+cd+da}$$ Let $S_1=\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}+\sqrt{ac}+\sqrt{bd},S_2=ab+bc+cd+da$ Now we need to prove: $$\frac{4+2S_1}{4+S_2}\ge2\implies S_1-S_2\ge2$$ If we prove $S_1\ge6$ and $S_2\le4$, then $S_1-S_2\ge6-4=2$. Both inequalities can be proved easily and both hold when $a=b=c=d=1$: $$(a+b+c+d)(b+c+d+a)\ge(ab+bc+cd+da)^2\implies S_2^2\le16\\\text{ since }a,b,c,d>0\implies S_2\le4$$

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  • $\begingroup$ @Integrator yup $\endgroup$ – RE60K Dec 7 '14 at 11:10
  • $\begingroup$ Hello,I think $S_{1}\ge 6$ it's not true such as $a=2,b=1,c=0.5,d=0.5$ see wolf give use $\approx 5.32\cdots$,see wolframalpha.com/input/?i=sqrt%282*1%29%2Bsqrt%282*0.5%29%2Bsqrt%282*0.5%29%2Bsqrt%281*0.5%29%2Bsqrt%281*0.5%29%2Bsqrt%280.5*0.5%29 $\endgroup$ – china math Dec 7 '14 at 15:24
  • $\begingroup$ @chinamath actually the inequality i proposed is the reverse way out i.e. S1<=6 $\endgroup$ – RE60K Dec 7 '14 at 15:27
  • $\begingroup$ and $S_{1}\le 6$,then we can't have $S_{1}-S_{2}\ge 2$ alway is true,so I think $S_{1}-S_{2}\ge 2$ is not true $\endgroup$ – china math Dec 7 '14 at 15:42
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Let's denote the left-hand expression by L. Then, by Cauchy-Schwarz:

$[a(a + bc) + b(b + cd) + c(c + da) + d(d+ab)]L \geq (a + b + c + d)^2 = 16$

Now, we have to show

$[a(a + bc) + b(b + cd) + c(c + da) + d(d+ab)] \leq 8$ and we are over. Let's write it down like: $A(a, b, c, d) = a^2 + b^2 + c^2 + d^2 + ac(b+d) + bd(a+c) \leq 8$

Notice that we get equality for $a = c, b = d$. I'll try this tactic: if I prove that $A(a, b, c, d) \leq A(\frac{a + c}{2}, b, \frac{a+c}{2}, d)$ and $A(a, b, c, d) \leq A(a, \frac{b + d}{2}, c, \frac{b + d}{2})$ (these two statements are analogous, so I'll just try to show the first) we will have this chain:

$A(a, b, c, d) \leq A(\frac{a + c}{2}, b, \frac{a+c}{2}, d) \leq A(\frac{a + c}{2}, \frac{b + d}{2}, \frac{a+c}{2}, \frac{b + d}{2}) = 8$.

Let's evaluate:

$A(\frac{a + c}{2}, b, \frac{a+c}{2}, d) - A(a, b, c, d) = \\ = 2(\frac{a + c}{2})^2 - a^2 - c^2 +(b+d)[(\frac{a + c}{2})^2 - ac] \\ = -2(\frac{a - c}{2})^2 + [4 - (a+c)](\frac{a - c}{2})^2 \\ = (\frac{a - c}{2})^2[2 - (a+c)]$

This works just if $a + c \leq 2$. If it's not true, $b + d \leq 2$ is, so we can start with the other substitution, and make at least one step in our chain. So, we can assume $a + c \leq 2$ and now we deal with:

$2a^2 + b^2 + d^2 + a^2(b + d) + 2abd \leq 8$

where $2a + b + d = 4, a \leq 1$.

In the next few calculations, I'm just using AM-GM on $bd$ and $b + d = 4 - 2a$.

$2a^2 + b^2 + d^2 + a^2(b + d) + 2abd \\ = 2a^2 + (b+d)^2 + a^2(b + d) + 2(a-1)bd \\ = 2a^2 + 4(2-a)^2 + 2a^2(2 - a) + 2(a - 1)bd \\ \leq 2a^2 + 4(2-a)^2 + 2a^2(2 - a) + 2(a - 1)(2-a)^2$

After tidying this up a little bit, it is easy to see that for $a = 0$ and $a = 1$ upper expression equals 8. Differentiate it twice to show that it's convex, and we are finished.

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  • $\begingroup$ Your second inequality is wrong. Try $a=3$,$b=1$ and $c=d=0.$ $\endgroup$ – Michael Rozenberg Jul 23 '19 at 16:46
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Take $1^{st}$ and $3^{rd}$ term apply the inequality, then take $2^{nd}$ and $4^{th}$ term and apply the inequality, you get

$$\ge 2\left(\frac1{1+\sqrt{bd}}+\frac1{1+\sqrt{ac}}\right)$$

again apply the inequality you get $\displaystyle \ge 2\left(\frac2{1+\sqrt{abcd}}\right)$

for $abcd =1$ you get your equality ie $a=b=c=d=1$

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    $\begingroup$ Are you sure $$\sqrt{bd}\ge 1,\sqrt{ac}\ge 1$$ and $$\sqrt{abcd}\ge 1?$$ In fact $abcd\le 1$..so I think you methods is wrong $\endgroup$ – china math Jun 18 '14 at 1:37
  • $\begingroup$ As @chinamath points out, the referred inequality does not give you the desired result directly, you need to address this. $\endgroup$ – Macavity Jun 18 '14 at 12:14
  • $\begingroup$ @Macavity,yes, so How prove it? Thank you $\endgroup$ – china math Jun 18 '14 at 12:23
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There is another method in which we can show that your inequality is true. It would certainly not start with your first condition and proving the result.The logic is other way round.It is that if there exists that inequality then we will prove or false it. I remember most of my graduate subject of metric spaces theorems were proved that way. There is a fact in derivatives that if $f(x)>0$ then function is increasing and $f'(x)>0$.

By making your inequality a function(your case is montonic increasing) of variables $a,b,c,d$(taking $2$ to the left hand side) and taking their partial derivatives w.r.t $a,b,c,$ and $d$ and checking whether each of them is positive we can deduce that the function is increasing.

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Let $a=\min\{a,b,c,d\}$, $b=a+u,$ $c=a+v$ and $d=a+w$.

Thus, $u$, $v$ and $w$ are non-negatives and we need to prove that $$\sum_{cyc}a\sum_{cyc}\frac{a}{a(a+b+c+d)+4bc}\geq2$$ or $$64(3(u^2+v^2+w^2)-2(uv+uw+vw))a^6+$$ $$+16(11(u^3+v^3+w^3)+21u^2v+13u^2w+5v^2u+21v^2w+13w^2u+5w^2v-50uvw)a^5+$$ $$+(43(u^4+v^4+w^4)+380u^3v+204u^3w+124v^3u+380v^3w+204w^3u+124w^3v+290u^2v^2+322u^2w^2+290v^2w^2+148u^2vw-168v^2uw-364w^2uv)a^4+$$ $$+(5(u^5+v^5+w^5)+85u^4v+33u^4w+29v^4u+85v^4w+33w^4u+29w^4v+270u^3v^2+138u^3w^2+86v^3u^2+270v^3w^2+138w^3u^2+86w^3v^2+420u^3vw+220v^3uw-12w^3uv+234u^2v^2w+230u^2w^2v-142v^2w^2u)a^3+$$ $$+(7u^5v+3u^5w+5v^5u+7v^5w+3w^5u+5w^5v+50u^4v^2+12u^4w^2+14v^4u^2+50v^4w^2+12w^4u^2+14w^4v^2+52u^3v^3+18u^3w^3+52v^3w^3+73u^4vw+47v^4uw-13w^4uv+314u^3v^2w+198u^3w^2v+134v^3u^2w+140v^3w^2u+114w^3u^2v-78w^3v^2u+140u^2v^2w^2)a^2+$$ $$+(2u^5v+5u^5w+w^5u+2w^5v+6u^4v^2+16u^4w^2+2v^4u^2+2v^4w^2+8w^4u^2+6w^4v^2+6u^3v^3+18u^3w^3+6v^3w^3+50u^4vw+5v^4uw-18w^4uv+58u^3v^2w+104u^3w^2v+18v^3u^2w+22v^3w^2u+36w^3u^2v-2w^3v^2u+76u^2v^2w^2)va+2u^2v^2w(u+v+w)^3\geq0,$$ which is obviously true by AM-GM.

Buffalo Way forever!

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