shobber wasn't this problem already posted?since i can't find the link right now i will tell the main ideas of two solutions:one posted here(you can also find it in old and new inequalities since this is vasc's problem posted in "gazeta matematica")and one that i found. one elegant idea would be to use the fact that $1/(a+1)^2 +1/(b+1)^2\ge 1/(ab+1)$ however if you are not inspired enough to see that you may also try to kill it with sturm and a little "strategy" since from sturm you get that $1/(a+1)^2 +1/(b+1)^2$is decreasing when $a-b$ decreases(when assuming that a>b)if $ab\ge1$.so you always pick two numbers such that their product is bigger than one.if anybody wants to know more about this solution i will post it

yep. it has been posted before.... http://www.artofproblemsolving.com/Forum/viewtopic.php?p=112435#p112435 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=190407#p190407

bodom wrote: shobber wasn't this problem already posted?since i can't find the link right now i will tell the main ideas of two solutions:one posted here(you can also find it in old and new inequalities since this is vasc's problem posted in "gazeta matematica")and one that i found. one elegant idea would be to use the fact that $1/(a+1)^2 +1/(b+1)^2\ge 1/(ab+1)$ however if you are not inspired enough to see that you may also try to kill it with sturm and a little "strategy" since from sturm you get that $1/(a+1)^2 +1/(b+1)^2$is decreasing when $a-b$ decreases(when assuming that a>b)if $ab\ge1$.so you always pick two numbers such that their product is bigger than one.if anybody wants to know more about this solution i will post it Can you post your solution bodom?

ok.i will do that in the evening because now i'm a little busy

shobber wrote: Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \] $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq\min\{1,\frac{2}{1+abcd}\}$ for all non-negative $a,$ $b,$ $c$ and $d.$ It is stronger.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=98967 http://www.mathlinks.ro/Forum/viewtopic.php?t=66028 http://www.mathlinks.ro/Forum/viewtopic.php?t=83784 http://www.mathlinks.ro/Forum/viewtopic.php?t=32031 http://www.mathlinks.ro/Forum/viewtopic.php?t=30724 http://www.mathlinks.ro/Forum/viewtopic.php?t=16027

shobber wrote: Let $ a,b,c,d > 0$ and $ abcd = 1$. Prove that $ \frac {1}{(1 + a)^2} + \frac {1}{(1 + b)^2} + \frac {1}{(1 + c)^2} + \frac {1}{(1 + d)^2} \geq 1$. Note $ f(r) = \frac {(1 + r)^2}{(1 + ra)^2} + \frac {(1 + r)^2}{(1 + rb)^2} + \frac {(1 + r)^2}{(1 + rc)^2} + \frac {(1 + r)^2}{(1 + rd)^2}$, the above iequality can be rewritten as $ f(1)\geq f(0)$. Further, the inequality $ f(p)\leq f(q)$ holds if and only if $ \left(0\leq p\leq q \wedge q \geq 1\right)\vee\left(p\leq q \leq 1 \wedge F(p,q)\geq 0\right)$, where $ F(p,q) = 791297088768p^{41}q^{41}(p + q)$ $ + 13824p^{40}q^{40}(427223039p^2 + 968927717pq + 427223039q^2)$ $ + 3456p^{39}q^{39}(p + q)(6090377113p^2 + 15189393892pq + 6090377113q^2)$ $ + 3456p^{38}q^{38}(13722903675p^4 + 64516122158p^3q + 100768147674p^2q^2$ $ + 64516122158pq^3 + 13722903675q^4)$ $ + 432p^{37}q^{37}(p + q)(173688858761p^4 + 838555981116p^3q + 1344852329190p^2q^2$ $ + 838555981116pq^3 + 173688858761q^4)$ $ \cdots$ $ - 162(135092061p^{16} - 665629194p^{15}q + 249716562p^{14}q^2 - 2231356257p^{13}q^3$ $ + 6400603379p^{12}q^4 + 15107734327p^{11}q^5 + 42943235426p^{10}q^6 + 23849200372p^9q^7$ $ + 71842248824p^8q^8 + 23849200372p^7q^9 + 42943235426p^6q^{10} + 15107734327p^5q^{11}$ $ + 6400603379p^4q^{12} - 2231356257p^3q^{13} + 249716562p^2q^{14} - 665629194pq^{15}$ $ + 135092061q^{16})$ $ - 209952(p + q)^{12}$ is an irreducible polynomial (degree = 83, 1770 expanding terms, 15 print pages). Specially, $ F(0,1) = 0$; $ F\left(\frac {1}{4},\frac {1}{3}\right) = \frac {316365130337792711401324387061897099707046593}{32294785429008247205187499988123494907904}$; $ F\left(\frac {1}{5},\frac {1}{3}\right) = - \frac {46735935121369674620811954200789126517628587200217088}{2764333080235003035568297491408884525299072265625}$; $ F\left(\frac {1}{7},\frac {1}{2}\right) = \frac {2219213864953238297522670210813327836751708984375}{20936185994239763297898479586227222883598336}$; $ F\left(\frac {1}{8},\frac {1}{2}\right) = - \frac {14061357910842776796378414886758065820716927789808483}{730750818665451459101842416358141509827966271488}$. Hence, $ f(r)$ rise monotonously on $ [s, + \infty)$, where $ s = 0.28783\cdots$ be a root of the following irreducible polynomial $ 114481639s^{16} + 564075766s^{15} + 1183750488s^{14} + 1376656902s^{13} + 742228566s^{12}$ $ - 999306606s^{11} - 3588048396s^{10} - 5291401734s^9 - 4791695109s^8 - 3224090940s^7$ $ - 1993194540s^6 - 1007735328s^5 - 187439320s^4 + 105691796s^3 + 43664832s^2$ $ - 3297024s + 62208$.

The inequality is clearly equivalent to $\frac16 \sum_{sym}a^2b^2c^2+\sum_{sym}a^2b^2c+\frac12\sum_{sym}a^2b^2+2\sum_{sym}a^2bc+2\sum_{sym}a^2b+\frac43 \sum_{sym}abc + \frac12 \sum_{sym}a^2+2\sum_{sym}ab+\sum_{sym}a+\frac16\sum_{sym}1 \geq \frac1{24}\sum_{sym}a^2b^2c^2d^2+\frac13\sum_{sym}a^2b^2c^2d+\frac16\sum_{sym}a^2b^2c^2+\sum_{sym}a^2b^2cd+\sum_{sym}a^2b^2c+\frac43\sum_{sym}a^2bcd+\frac14\sum_{sym}a^2b^2+2\sum_{sym}a^2bc+\frac23\sum_{sym}abcd+\sum_{sym}a^2b+\frac43\sum_{sym}abc+\frac16\sum_{sym}a^2+\sum_{sym}ab+\frac13\sum_{sym}a+\frac1{24}\sum_{sym}1$ by expansion. After homogenising to degree 8 ${\frac16 \sum_{sym}a^{5/2}b^{5/2}c^{5/2}d^{1/2}+\sum_{sym}a^{11/4}b^{11/4}c^{7/4}d^{3/4}+\frac12\sum_{sym}a^3b^3cd+2\sum_{sym}a^3b^2c^2d+2\sum_{sym}a^{13/4}b^{9/4}c^{5/4}d^{5/4}+\frac43 \sum_{sym}a^{9/4}b^{9/4}c^{9/4}d^{5/4} + \frac12 \sum_{sym}a^{7/2}b^{3/2}c^{3/2}d^{3/2}+2\sum_{sym}a^{5/2}b^{5/2}c^{3/2}d^{3/2}+\sum_{sym}a^{11/4}b^{7/4}c^{7/4}d^{7/4}+\frac16\sum_{sym}a^2b^2c^2d^2 \geq \frac1{24}\sum_{sym}a^2b^2c^2d^2+\frac13\sum_{sym}a^{9/4}b^{9/4}c^{9/4}d^{5/4}+\frac16\sum_{sym}a^{5/2}b^{5/2}c^{5/2}d^{1/2}+\sum_{sym}a^{5/2}b^{5/2}c^{3/2}d^{3/2}+\sum_{sym}a^{11/4}b^{11/4}c^{7/4}d^{3/4}+\frac43\sum_{sym}a^{11/4}b^{7/4}c^{7/4}d^{7/4}+\frac14\sum_{sym}a^3b^3cd+2\sum_{sym}a^3b^2c^2d+\frac23\sum_{sym}a^2b^2c^2d^2+\sum_{sym}a^{13/4}b^9/4}c^{5/4}d^{5/4}+\frac43\sum_{sym}a^{9/4}b^{9/4}c^{9/4}d^{5/4}+\frac16\sum_{sym}a^{7/2}b^{3/2}c^{3/2}d^{3/2}+\sum_{sym}a^{5/2}b^{5/2}c^{3/2}d^{3/2}+\frac13\sum_{sym}a^{11/4}b^{7/4}c^{7/4}d^{7/4}+\frac1{24}\sum_{sym}a^2b^2c^2d^2$ which after simplification is conveniently equivalent to $\frac14\sum_{sym}a^3b^3cd+\sum_{sym}a^{13/4}b^{9/4}c^{5/4}d^{1/4}+\frac13\sum_{sym}a^{7/2}b^{3/2}c^{3/2}d^{3/2} \geq \frac13 \sum_{sym} a^{9/4}b^{9/4}c^{9/4}d^{5/4}+\frac23 \sum_{sym}a^{11/4}b^{7/4}c^{7/4}d^{7/4}+\frac7{12}\sum_{sym}a^2b^2c^2d^2$ But this is true since by Muirhead's inequality or weighted AM-GM we have $\frac14\sum_{sym}a^3b^3cd \geq \frac14\sum_{sym}a^{9/4}b^{9/4}c^{9/4}d^{5/4}$ $\frac34\sum_{sym}a^{13/4}b^{9/4}c^{5/4}d^{5/4} \geq \frac{1}{12}\sum_{sym}a^{9/4}b^{9/4}c^{9/4}d^{5/4}+\frac23 \sum_{sym}a^{11/4}b^{7/4}c^{7/4}d^{7/4}$ $\frac13\sum_{sym}a^{7/2}b^{3/2}c^{3/2}d^{3/2}+\frac14 \sum_{sym}a^{13/4}b^{9/4}c^{5/4}d^{5/4} \geq \frac{7}{12}\sum_{sym}a^2b^2c^2d^2$.

These bashes are really ugly!

For prove the problem , we will prove that : \[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \] \[ \frac {1}{(1+c)^2} + \frac {1}{(1+d)^2} \ge \frac {1}{1+cd} \] Proof : \[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \] \[ \Longleftrightarrow (1+ab)((1+a)^2+(1+b)^2) \ge (1+a)^2(1+b)^2 \] \[ \Longleftrightarrow a^3b+b^3a-a^2b^2-2ab+1 \ge 0 \] \[ \Longleftrightarrow 1+ab(a^2+b^2-ab-2) \ge 0 \] Use AM-GM on $LHS$ we have \[ \text {LHS} \ge 1+ab(2ab-ab-2) = 1+ab(ab-2)\1+a^2b^2-2ab \ge 2ab - 2ab = 0 \] Similarly we can prove second one. Now use $abcd=1$. So we have : \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \ge \frac {1}{ab+1} + \frac {1}{cd+1} \] \[ = \frac {cd+ab+2}{cd+ab+abcd+1} = \frac {cd+ab+2}{cd+ab+2} = 1 \]

The following inequality is also true. Let $a,b,c >0$ and $abc=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\]

Just take $d=1$ in the general case: $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq 1 $ You get \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{4}\geq 1 \] So: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq \frac{3}{4} \]

Let a,b,c> 0 abc=1 Using the previous lemma $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{1}{1+ab}=\frac{c}{c+1}$ We only need to prove that : $\frac{1}{(1+c)^2} + \frac{c}{c+1} \geq\frac{3}{4}$ which's equivalente after expanding to $( c-1)^2\geq0$

Thanks. Let $x_1,x_2,\cdots,x_n>0$ and $x_1x_2\cdots x_n=1$. Prove that: \[ \frac{1}{(1+x_1)^2}+\frac{1}{(1+x_2)^2}+\cdots+\frac{1}{(1+x_n)^2} \geq \frac{n}{4}. \]

see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=542038&p=3124726#p3124726

leminscate wrote: These bashes are really ugly! No, they're not.

sqing wrote: The following inequality is also true. Let $a,b,c >0$ and $abc=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\] Let $a=\frac{y}{x},b=\frac{z}{y},c= \frac{x}{z},$ we have (《Mathematical communication》(China Wuhan) N0.7、8(2014) Q183 ): Let $x,y,z$ be positive real numbers . Prove that \[ \frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}+\frac{z^2}{(z+x)^2}\geq\frac{3}{4}.\]

I think from the first we must use CS inequality, then Muirhead's inequality.

The following inequalities are also true. Let $a,b,c >0$ and $abc=1$. Prove that: \[\frac{1}{(a+1)^2} + \frac{1}{(b+1)^2} + \frac{1}{(c+1)^2} + \frac{1}{{a + b + c +1}} \geqslant 1.\] Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(3a-1)^2}+\frac{1}{(3b-1)^2}+\frac{1}{(3c-1)^2}+\frac{1}{(3d-1)^2} \geq 1. \] here

we can prove: $ f(a,b,c,d)\geq f(\sqrt{ab},\sqrt{ab},c,d)\geq 0 $ with $ f(a,b,c,d)=\sum (\frac{1}{(1+a)^{2}}-\frac{1}{4}) $

sqing wrote: The following inequality is also true. Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(3a-1)^2}+\frac{1}{(3b-1)^2}+\frac{1}{(3c-1)^2}+\frac{1}{(3d-1)^2} \geq 1. \] It's Jensen!

War-Hammer wrote: Proof : \[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \] \[ \Longleftrightarrow (1+ab)((1+a)^2+(1+b)^2) \ge (1+a)^2(1+b)^2 \] \[ \Longleftrightarrow a^3b+b^3a-a^2b^2-2ab+1 \ge 0 \] \[ \Longleftrightarrow 1+ab(a^2+b^2-ab-2) \ge 0 \] which is $ab(a-b)^2+(ab-1)^2\geq0$.

generalization to 6 variables: let $x_1,x_2,\dots ,x_6$ be 6 positive numbers such that $x_1x_2\dots x_6=1$ and $\prod_{1\le i<j\le 6}(x_ix_j-1)\ge 0$. then $$\sum_{i=1}^{6}\frac{1}{(x_i+1)^2}\ge \frac{3}{2}$$

Crazy long bashes

From Cauchy Schwarz, $(x+y)(1+xy)\ge (\sqrt{y}+\sqrt{xy})^2\Leftrightarrow \dfrac{1}{(1+x)^2}\ge \dfrac{y}{x+y}\cdot \dfrac{1}{1+xy}$, hence $\dfrac{1}{(1+x)^2}+\dfrac{1}{(1+y)^2}\ge \dfrac{1}{1+xy}$. Thereby, $\dfrac{1}{(1+a)^2}+\dfrac{1}{(1+b)^2}+\dfrac{1}{(1+c)^2}+\dfrac{1}{(1+d)^2}\ge \dfrac{1}{1+ab}+\dfrac{1}{1+cd}=1$

$(x+y)(1+xy)\geq (\sqrt{y}+x\sqrt{y})^2 \iff (x+y)(1+xy)\geq y(1+x)^2$

sqing wrote: sqing wrote: The following inequality is also true. Let $a,b,c >0$ and $abc=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\] Let $a=\frac{y}{x},b=\frac{z}{y},c= \frac{x}{z},$ we have (《Mathematical communication》(China Wuhan) N0.7、8(2014) Q183 ): Let $x,y,z$ be positive real numbers . Prove that \[ \frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}+\frac{z^2}{(z+x)^2}\geq\frac{3}{4}.\] Just is my LBQ105. BQ

sqing wrote: The following inequality is also true. Let $a,b,c >0$ and $abc=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\] Also substituting $ a=xy/z^2$ and likewise does the trick.

War-Hammer wrote: For prove the problem , we will prove that : \[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \]\[ \frac {1}{(1+c)^2} + \frac {1}{(1+d)^2} \ge \frac {1}{1+cd} \] Proof : \[ \frac {1}{(1+a)^2} + \frac {1}{(1+b)^2} \ge \frac {1}{1+ab} \]\[ \Longleftrightarrow (1+ab)((1+a)^2+(1+b)^2) \ge (1+a)^2(1+b)^2 \]\[ \Longleftrightarrow a^3b+b^3a-a^2b^2-2ab+1 \ge 0 \]\[ \Longleftrightarrow 1+ab(a^2+b^2-ab-2) \ge 0 \] Use AM-GM on $LHS$ we have \[ \text {LHS} \ge 1+ab(2ab-ab-2) = 1+ab(ab-2)\1+a^2b^2-2ab \ge 2ab - 2ab = 0 \] Similarly we can prove second one. Now use $abcd=1$. So we have : \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \ge \frac {1}{ab+1} + \frac {1}{cd+1} \]\[ = \frac {cd+ab+2}{cd+ab+abcd+1} = \frac {cd+ab+2}{cd+ab+2} = 1 \] thank you for Lemma

Let $a,b,c,d$ are positive reals such that $abcd = 1 .$ Prove that$$\frac{1}{5a^2-2a+1} +\frac{1}{5b^2-2b+1} + \frac{1}{5c^2-2c+1} + \frac{1}{5d^2-2d+1} \geq \frac{1}{(a^2+1)^2} +\frac{1}{(b^2+1)^2} +\frac{1}{(c^2+1)^2} +\frac{1}{(d^2+1)^2} \geq 1.$$

shobber wrote: Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \] here here here:

Let $a,b,c >0$ and $abc=1$. Prove that $$\dfrac{1}{(3a^2+1)^2}+\dfrac{1}{(3b^2+1)^2}+\dfrac{1}{(3c^2+1)^2}+\dfrac{1}{(a+b+c+1)^2}\geq\dfrac{1}{4}$$ shobber wrote: Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \] Poland 2018, MO Science Camp h Romanian magazine Gazeta Matematica, 1999, No. 11

sqing wrote: The following inequality is also true. Let $a,b,c >0$ and $abc=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\geq\frac{3}{4}.\] Let $a,b,c\in(0,2)$ and $abc=(1-a)(2-b)(3-c).$ Prove that$$\dfrac{a^2}{3}+\dfrac{b^2}{12}+\dfrac{c^2}{27}\geq\frac{3}{4}.$$

Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}<3 \]

sqing wrote: Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}<3 \] Because $$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq1.$$

sqing wrote: Let $a,b,c >0$ and $abc=1$. Prove that: \[\frac{1}{(a+1)^2} + \frac{1}{(b+1)^2} + \frac{1}{(c+1)^2} + \frac{1}{{a + b + c +1}} \geqslant 1.\]

arqady wrote: shobber wrote: Let $a,b,c,d >0$ and $abcd=1$. Prove that: \[ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1 \] $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq\min\{1,\frac{2}{1+abcd}\}$ for all non-negative $a,$ $b,$ $c$ and $d.$ It is stronger. Let us not forget this math problem

This problem is a special case of this inequality: Let $abcd=1$ as above. Then $\frac{1}{(1+a)^k}+\frac{1}{(1+b)^k}+\frac{1}{(1+c)^k}+\frac{1}{(1+d)^k} \geq 2^{2-k}$.

Very interesting

sqing wrote: Thanks. Let $x_1,x_2,\cdots,x_n>0$ and $x_1x_2\cdots x_n=1$. Prove that: \[ \frac{1}{(1+x_1)^2}+\frac{1}{(1+x_2)^2}+\cdots+\frac{1}{(1+x_n)^2} \geq \frac{n}{4}. \] This is wrong for $n\geq 5$. Try $x_1=...=x_{n-1}\to +\infty$.