bgn wrote: Let $a,b,c,d$ be positive real numbers with $a+b+c+d=2$. Prove the inequality: $$\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right )$$ Very nice . Very beautiful.

bgn wrote: Let $a,b,c,d$ be positive real numbers with $a+b+c+d=2$. Prove the inequality: $$\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right )$$ $\frac{4}{(a+b)(c+d)}+4-4\left(\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1}\right)$ $=\frac{3(a+b-c-d)^2}{4(a+b)(c+d)(a+b+1)(c+d+1)}\geq 0$ $\therefore \frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq \frac{(a+b+c+d)^2}{(a+b)(c+d)}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right )$

$\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1}=(\frac{a+b+1}{c+d+1}+1)+(\frac{c+d+1}{a+b+1}+1)-2=\frac{16}{(a+b+1)(c+d+1)}-2$ $\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+12 \geq \frac{64}{(a+b+1)(c+d+1)}$ $(\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+12)((ab+bc)+(ac+bd)+3)\geq 64$ (cauchy) and $(ab+bc)+(ac+bd)+3=(a+b+1)(c+d+1)$ We are done.

MonsterS wrote: $\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1}=(\frac{a+b+1}{c+d+1}+1)+(\frac{c+d+1}{a+b+1}+1)-2=\frac{16}{(a+b+1)(c+d+1)}-2$ $\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+12 \geq \frac{64}{(a+b+1)(c+d+1)}$ $(\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+12)((ab+bc)+(ac+bd)+3)\geq 64$ (cauchy) and $(ab+bc)+(ac+bd)+3=(a+b+1)(c+d+1)$ We are done. Nice. $$\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1}+2=\frac{16}{(a+b+1)(c+d+1)}$$

The following stronger inequality is also true: Let $a+b+c+d=4$. Show that: $$\frac{(a+c)^2}{ad+bc}+\frac{(b+d)^2}{ac+bd}+\frac{1}{2} \geq \frac{9}{4}\left(\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right)$$

I think its easy for iran

Using titu inequality this can be simplified to $\frac{1}{(a+b)(c+d)}\geq\frac{a+b+1}{c+d+1} +\frac{c+d+1}{a+b+1}$ Now let $a+b=x,$$c+d=y$ $then$ $again$ $after$ $simplification $ $(xy)^2-2xy+1\geq 0$ Which is equivalent to $(xy-1)^2\geq 0$ Which is true $Q.E.D$

enhanced wrote: Using titu inequality this can be simplified to $\frac{1}{(a+b)(c+d)}\geq\frac{a+b+1}{c+d+1} +\frac{c+d+1}{a+b+1}$ I think it should be $\frac{1}{(a+b)(c+d)}+1\geq\frac{a+b+1}{c+d+1} +\frac{c+d+1}{a+b+1}$, but it still works.

My solution:First note that by caushy we have: $(\frac{(a+c)^2}{ad+bc} +\frac{(b+d)^2}{ac+bd}+12)(ad+bc+ac+bd+3) \ge 64$ Also note that $ad+bc+ac+bd+3=(a+b+1)(c+d+1)$ and also $64=((a+b+1)+(c+d+1))^2$.Now dealing with some calculations give the desires inequality.

The right hand side of the desired inequality is equivalent to, $$4\left(\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1}\right) = 4\left(\frac{(a+b+1)^2+(c+d+1)^2}{(a+b+1)(c+d+1)}\right) $$ $$ =4\left( \frac{ (a+b)^2+2(a+b)+1 + (c+d)^2+2(c+d)+1}{(a+b)(c+d) +(a+b)+(c+d)+1}\right)$$ Using the condition $a+b+c+d=2$, this is equal to $$ =4\left( \frac{ (a+b)^2+(c+d)^2+3(a+b+c+d)}{(a+b)(c+d) +\frac{3}{2}(a+b+c+d)}\right)$$ Again using the condition $a+b+c+d=2$, we obtain the fraction in the homogenized form, $$= 4\left( \frac{ (a+b)^2+(c+d)^2+\frac{3(a+b+c+d)^2}{2}}{(a+b)(c+d) +\frac{3}{4}(a+b+c+d)^2}\right)$$ $$ =4\left( \frac{4 (a+b)^2+4(c+d)^2+6(a+b+c+d)^2}{4(a+b)(c+d) +3(a+b+c+d)^2}\right)$$ $$ =4\left( \frac{4 (a+b)^2+4(c+d)^2+6(a+b)^2+6(c+d)^2+12(a+b)(c+d)}{4(a+b)(c+d) +3(a+b)^2+3(c+d)^2+6(a+b)(c+d)}\right)$$ $$ = 4\left( \frac{10 (a+b)^2+10(c+d)^2+12(a+b)(c+d)}{3(a+b)^2+3(c+d)^2+10(a+b)(c+d)}\right) \quad \quad \quad \quad \quad \quad (1)$$ Now, using Cauchy-Schwarz inequality, we obtain $$\frac{(a+c)^2}{ad+bc}+\frac{(b+d)^2}{ac+bd} \enspace + \enspace 4 \enspace \geq \enspace \frac{(a+c+b+d)^2}{ad+bc+ac+bd} \enspace + \enspace 4 \enspace = \enspace\frac{(a+b)^2+(c+d)^2+2(a+b)(c+d)}{(a+b)(c+d)} \enspace + \enspace 4 $$ $$\frac{(a+b)^2+(c+d)^2+2(a+b)(c+d)}{(a+b)(c+d)} \enspace + \enspace 4 \enspace = \enspace \frac{(a+b)^2+(c+d)^2+6(a+b)(c+d)}{(a+b)(c+d)} \quad \quad \quad \quad \quad \quad (2)$$ From $(1)$ and $(2)$, we observe that to prove the desired inequality we just need to show that, $$ \frac{(a+b)^2+(c+d)^2+6(a+b)(c+d)}{(a+b)(c+d)} \enspace \geq \enspace 4\left( \frac{10 (a+b)^2+10(c+d)^2+12(a+b)(c+d)}{3(a+b)^2+3(c+d)^2+10(a+b)(c+d)}\right) $$ Now, let $x=(a+b)$ and $y = (c+d)$, so the last inequality is equivalent to, $$ \frac{x^2+y^2+6xy}{xy} \enspace \geq \enspace 4\left( \frac{10x^2+10y^2+12xy}{3x^2+3y^2+10xy}\right)$$ $$ \Leftrightarrow 3x^4+3y^4+66x^2y^2+28x^3y+28xy^3 \enspace \geq \enspace 40x^3y+40xy^3+48x^2y^2$$ $$ \Leftrightarrow 3x^4+ 3y^4+18x^2y^2-12x^3y-12xy^3 \geq 0$$ $$ \Leftrightarrow 3 ( x^4-4x^3y +6x^2y^2 -4xy^3+y^4) \geq 0$$ $$ \Leftrightarrow 3(x-y)^4 \geq 0$$ which is obviously true. Equality occurs iff $x=y \Leftrightarrow a+b=c+d=1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \blacksquare$

By Titu $$\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd} \geq \frac{(a+b+c+d)^2}{(a+b) (c+d)}=\frac{4}{(a+b) (c+d)}$$Let $x=a+b$, $y=c+d$, then $x+y=2$ and it suffices to show that $$\frac{1}{xy}+1 \geq \frac{x+1}{y+1}+\frac{y+1}{x+1}$$$$(1+xy)(1+x)(1+y) \geq xy((1+x)^2 + (1+y)^2)$$Substituting $y=2-x$ we need to prove $$(x^2-2x-1)(x^2-2x-3) \geq -2(x^2-2x)(x^2-2x+5)$$Let $t=x^2-2x$. Then the desired inequality is $$(t-1)(t-3) \geq -2t(t+5)$$$$3(t+1)^2 \geq 0$$Which is obviously true.

Note that it suffices to show that : $\frac{(a+c)^2}{ad+bc}+\frac{(b+d)^2}{ac+bd}+12 \ge \frac{64}{3+(a+b)(c+d)}$ wich can be trivially proved by C.S

bgn wrote: Let $a,b,c,d$ be positive real numbers with $a+b+c+d=2$. Prove the following inequality: $$\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right).$$ Proposed by Mohammad Jafari here

Let $(a+b)(c+d)=p$ then by T2's lemma, $$\frac{(a+c)^2}{ad+bc}+\frac{(b+d)^2}{ac+bd}\geq \frac{(a+b+c+d)^2}{ad+bc+ac+bd}=\frac{4}{p}$$The other side simplifies to $$\frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1}=\frac{((a+b+1)+(c+d+1))^2}{(a+b+1)(c+d+1)}-2=\frac{16}{p+3}-2=\frac{10-2p}{p+3}$$Thus it suffices to show that $$\frac{4}{p}+4\geq 4\left(\frac{10-2p}{p+3}\right)\iff \frac{p+1}{p}\geq \frac{10-2p}{p+3}\iff $$$$(p+1)(p+3)\geq p(10-2p)\iff 3p^2+3\geq 6p\iff (p-1)^2\geq 0$$