Prove $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ for positives $\sum a = 4$

Question

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 $a=3$

Answer 1

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 $a=b=c=d=1$: $$(a+b+c+d)(b+c+d+a)\ge(ab+bc+cd+da)^2\implies S_2^2\le16\\\text{ since }a,b,c,d>0\implies S_2\le4$$

Answer 2

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.

Answer 3

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.

Answer 4

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!

Answer 5

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 $a=b=c=d=1$