For $a,\ b,\ c>0$, $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq 1\Longleftrightarrow abc\geq a+b+c+2$, yielding $c(ab-1)\geq a+b+2>0$, thus $ab>1$. $\therefore ab+bc+ca-12=ab+c(a+b)-12\geq ab+\frac{(a+b)(a+b+2)}{ab-1}-12$ $\geq ab+\frac{4ab+4\sqrt{ab}}{ab-1}-12$ $\geq \frac{(\sqrt{ab}+3)(\sqrt{ab}-2)^2}{\sqrt{ab}-1}\geq 0.$ The equality holds when $ab=4$, therefore $(a^2+1)(b^2+1)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2\geq 11^2+2^2=125.$

As kunny pointed out, the given condition is equivalent to $abc \ge a+b+c+2$ Also note that \[1\ge \sum\frac1{1+a} \ge \frac3{\sqrt[3]{(1+a)(1+b)(1+c)}} \implies (1+a)(1+b)(1+c) \ge 27\] Therefore \[(a+b+c+3)^3 \ge 27(a+1)(b+1)(c+1) \ge 3^6 \implies a+b+c \ge 6\] This implies $abc \ge 8$ Hence by Holder (one can also use AM-GM) \[(1+a^2)(1+b^2)(1+c^2) \ge (1+(abc)^{\frac23})^3 \ge 125\]

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My solution

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A second copy

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Sayan wrote: Therefore \[a+b+c \ge 6\] This implies $abc \ge 8$ Try $a=b=\frac{1}{2}$ and $c=5$.

arqady wrote: Sayan wrote: Therefore \[a+b+c \ge 6\] This implies $abc \ge 8$ Try $a=b=\frac{1}{2}$ and $c=5$. If you just quote that thing, it will be false. Of course, that implication was due to $abc\ge a+b+c+2$.

I understand you! It's very nice your solution, Sayan! By $uvw$'s substitutions we obtain $w^3\geq3u+2\geq3w+2$. Hence, $w^3\geq8$ and your Holder.

After kunny pointed out that $abc\geq a+b+c+2$ and after I noted from the given inequalities that $a+b+c\geq6$, we can see that $abc\geq 8$. $(1+a^2)(1+b^2)(1+c^2)= (1+\dfrac{a^2}{4}+\dfrac{a^2}{4}+\dfrac{a^2}{4}+\dfrac{a^2}{4})(1+\dfrac{b^2}{4}+\dfrac{b^2}{4}+\dfrac{b^2}{4}+\dfrac{b^2}{4})(1+\dfrac{c^2}{4}+\dfrac{c^2}{4}+\dfrac{c^2}{4}+\dfrac{c^2}{4})\geq 5^3(abc)^{\dfrac{8}{5}}/2^\dfrac{24}{5}=125$

I see a great deal of nice solutions, I will endeavour to add a different one: using the double-tangent method, I will prove that the following function is minimal at $x = 2$ and the minimum is 0. So consider the function: \[f(x) = ln(1+x^2)-ln(5) -\frac{36}{5}(\frac{1}{3}-\frac{1}{x+1})\] The first derivative is: \[f'(x) = \frac{2(x-2)(5x^2+2x+9)}{5(x+1)^2(x^2+1)}\] Then f is minimal at $x = 2$ and we have: \[f(x) \geq f(2) =0\] Thus: \[\sum{ln(1+a^2)}\geq \sum{(ln(5)+\frac{36}{5}(\frac{1}{3}-\frac{1}{a+1}))} \geq ln(125)+\frac{36}{5}(1-\sum{\frac{1}{a+1}})\] Which implies that: \[(1+a^2)(1+b^2)(1+c^2) \geq 125\]

Generalization Let $a_1,a_2,\cdots,a_n$ $(n\ge 2)$be positive real numbers such that$ \cfrac{1}{1+a_1}+\cfrac{1}{1+a_2}+\cdots+\cfrac{1}{1+a_n}\le 1. $Prove that \[(1+a_1^2)(1+a_2^2)\cdots (1+a_n^2)\ge \left(1+(n-1)^2\right)^n.\] n=4

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[/quote] sqing wrote: Generalization Let $a_1,a_2,\cdots,a_n$ $(n\ge 2)$be positive real numbers such that$ \cfrac{1}{1+a_1}+\cfrac{1}{1+a_2}+\cdots+\cfrac{1}{1+a_n}\le 1. $Prove that \[(1+a_1^2)(1+a_2^2)\cdots (1+a_n^2)\ge \left(1+(n-1)^2\right)^n.\] For the general case, I will just generalise the solution: I will prove that the following function is minimal at $x = n-1$ and the minimum is 0. So consider the function: \[f_{n}(x) = ln(1+x^2)-ln(1+(n-1)^2) -\frac{2n^2(n-1)}{1+(n-1)^2}(\frac{1}{n}-\frac{1}{x+1})\] The first derivative is: \[f'_{n}(x) = \frac{2(x-(n-1))((n^2-2n+2)x^2+2x+n^2)}{(n^2-2n+2)(x+1)^2(x^2+1)}\] Then f is minimal at $x = n-1$ and we have: \[f_{n}(x) \geq f_{n}(n-1) =0\] Thus: \[\sum{ln(1+a_{i}^2)}\geq \sum{(ln(1+(n-1)^2)+\frac{2n^2(n-1)}{1+(n-1)^2}(\frac{1}{n}-\frac{1}{a_{i}+1})) \geq ln(1+(n-1)^2)^n}\] Which implies that: \[(1+a_1^2)(1+a_2^2)\cdots (1+a_n^2)\ge \left(1+(n-1)^2\right)^n.\]

my proof = by the given condition we have $ abc >= a+b+c+2 $ now by titu's lemma , $ 1>= (1/1+a) + (1/1+b) + (1/1+c) >= (3^2 /3+a+b+c) = (9/3+a+b+c) $ and hence $ a+b+c >= 6 $ now the given inequality is equivalent to prove that $ a^2b^2c^2 + a^2+b^2+c^2 + 1 +a^2b^2 + a^2c^2 +b^2c^2 >= 125 $ since $ a^2b^2c^2 >= 64 $ and $ a^2+b^2+c^2 >= {(a+b+c)^2 /3} >= 12 $ thus it suffices to prove that $ a^2b^2 + b^2c^2 + a^2c^2 >= 48 $ which is indeed true by AM-GM , since $ a^2b^2+b^2c^2 + c^2a^2 >= 3 (abc)^{4/3} >= 3 ( 8)^{4/3} = 48 $ hence proved

YESMAths wrote: Let $a,b,c$ be positive real numbers such that \[ \cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1. \] Prove that $(1+a^2)(1+b^2)(1+c^2)\ge 125$. When does equality hold? Proof ：Let $x = \frac{1}{{1 + a}},y = \frac{1}{{1 + b}},z = \frac{1}{{1 + c}}, $,$x,y,z > 0,x + y + z \le 1$， \[abc = \left( {\frac{1}{x} - 1} \right)\left( {\frac{1}{y} - 1} \right)\left( {\frac{1}{z} - 1} \right)\] \[ = \frac{{(1 - x(1 - y)(1 - z)}}{{xyz}} \ge \frac{{(y + z)(z + x)(x + y)}}{{xyz}} \ge 8\] \[(1 + a^2 )(1 + b^2 )(1 + c^2 ) \ge \left[ {1 + \sqrt[3]{{\left( {abc} \right)^2 }}} \right]^3 \ge \left[ {1 + \sqrt[3]{{\left( 8 \right)^2 }}} \right]^3 = 125\] Generalization： Let $a_i > 0(i = 1,2, \cdots ,n),n \ge 3,m \in N,m \ge 1,$ such that \[\sum\limits_{i = 1}^n {\left( {\frac{1}{{1 + a_i }}} \right)} ^{\frac{1}{m}} = s \le 1,k > 0\] Prove that \[\prod\limits_{i = 1}^n {\left( {1 + a_i^k } \right)} \ge \left\{ {1 + \left[ {\left( {\frac{n}{s}} \right)^m - 1} \right]^k } \right\}^n \]

sqing wrote: Generalization Let $a_1,a_2,\cdots,a_n$ $(n\ge 2)$be positive real numbers such that$ \cfrac{1}{1+a_1}+\cfrac{1}{1+a_2}+\cdots+\cfrac{1}{1+a_n}\le 1. $Prove that \[(1+a_1^2)(1+a_2^2)\cdots (1+a_n^2)\ge \left(1+(n-1)^2\right)^n.\] n=4 Let $x_i=\frac{1}{{1 + a_i}}$ $(i=1,2,\cdots,n)$ , hence $\prod_{i=1}^{n}a_i = \frac{{\prod_{i=1}^{n}(1 - x_i)}}{{\prod_{i=1}^{n}x_i }}\ge $ $\frac{{(x_2+x_3+\cdots+x_n)(x_3+x_4+\cdots+x_n+x_1)\cdots(x_1+x_2+\cdots+x_{n-2}+x_{n-1})}}{{\prod_{i=1}^{n}x_i }}$ $\ge (n-1)^n$, By AM-GM, $\prod_{i=1}^{n}{1+a_i^2}\ge ((n-1)^2+1)^n\prod_{i=1}^{n}{\sqrt[(n-1)^2+1]{\frac{a_i^{2(n-1)^2}}{(n-1)^{2(n-1)^2}}}}\ge \left(1+(n-1)^2\right)^n.$

To compare your generalization with mine !

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we denote x_k/2015=a_k and get first generalization.

Let $a,b,c$ be positive real numbers such that $\cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1$ .Prove that\[(1+bc)(1+ca)(1+ab)\geq125. \]here

Sayan wrote: Hence by Holder (one can also use AM-GM) \[(1+a^2)(1+b^2)(1+c^2) \ge (1+(abc)^{\frac23})^3 \ge 125\] Wondering why straight up using $AM-GM$ on each term gives the wrong answer. \[(1+a^2)(1+b^2)(1+c^2) \ge 8abc \ge 64\]

What does the notation $\sum_{cyc}$ mean?

Also, another way to get $abc\geq 8$, $$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq 1$$Adding $\frac{a}{1+a}$ on both sides, gives $$\frac{a}{1+a} \geq \frac{1}{1+b}+\frac{1}{1+c}$$Similarly for $b$ and $c$. Now, $$\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{2}{\sqrt{(1+b)(1+c)}}$$$$\Rightarrow \frac{a}{1+a} \geq \frac{2}{\sqrt{(1+b)(1+c)}}$$Multiplying the similar inequalities for $b$ and $c$ gives $abc \geq 8$.

1/(1+a) + 1/(1+b) + 1/(1+c) <= 1 implies that abc>=2+a+b+c but from am-gm , 2+a+b+c>=2+3(abc)^(1/3) thus abc>=2+3(abc)^(1/3) substituting abc=t^3 , we get (t+1)(t+1)(t-2)>=0 but as a,b,c are +ve , so is t thus (t+1)^2 is always +ve thus t>=2 which implies that abc>=8 from am-gm a^2 + b^2 + c^2 >=3(abc)^(2/3)>=12 (ab)^2 + (bc)^2 + (ca)^2 >=3(abc)^(4/3)>=48 (abc)^2>=64 adding the above three inequations we get a^2 + b^2 + c^2 + (ab)^2 + (bc)^2 + (ca)^2 + (abc)^2>=124 adding 1 on both sides we get the desired result

equinox145111 wrote: What does the notation $\sum_{cyc}$ mean? Refer CYCLIC SUMMATION. NOTE: You should clear the basics before you use AOPS.(Polite Request)

andrew100 wrote: After kunny pointed out that $abc\geq a+b+c+2$ and after I noted from the given inequalities that $a+b+c\geq6$, we can see that $abc\geq 8$. $$(1+a^2)(1+b^2)(1+c^2)= (1+\frac{a^2}{4}+\frac{a^2}{4}+\frac{a^2}{4}+\frac{a^2}{4})(1+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4})(1+\frac{c^2}{4}+\frac{c^2}{4}+\frac{c^2}{4}+\frac{c^2}{4})\geq 5^3(abc)^{\frac{8}{5}}/2^\frac{24}{5}=125$$ Just fixed your latex..

Sayan wrote: As kunny pointed out, the given condition is equivalent to $abc \ge a+b+c+2$ Also note that \[1\ge \sum\frac1{1+a} \ge \frac3{\sqrt[3]{(1+a)(1+b)(1+c)}} \implies (1+a)(1+b)(1+c) \ge 27\]Therefore \[(a+b+c+3)^3 \ge 27(a+1)(b+1)(c+1) \ge 3^6 \implies a+b+c \ge 6\]This implies $abc \ge 8$ Hence by Holder (one can also use AM-GM) \[(1+a^2)(1+b^2)(1+c^2) \ge (1+(abc)^{\frac23})^3 \ge 125\] This solution is wrong since a + b + c >= 6 does not imply abc >= 8 (while the opposite of this is true, i.e. abc >= 8 implies a + b + c >= 6). As an counter-example, take a = 100, b = c = 1/10.

scpajmb wrote: YESMAths wrote: Let $a,b,c$ be positive real numbers such that \[ \cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1. \]Prove that $(1+a^2)(1+b^2)(1+c^2)\ge 125$. When does equality hold? Proof ：Let $x = \frac{1}{{1 + a}},y = \frac{1}{{1 + b}},z = \frac{1}{{1 + c}}, $,$x,y,z > 0,x + y + z \le 1$， \[abc = \left( {\frac{1}{x} - 1} \right)\left( {\frac{1}{y} - 1} \right)\left( {\frac{1}{z} - 1} \right)\]\[ = \frac{{(1 - x(1 - y)(1 - z)}}{{xyz}} \ge \frac{{(y + z)(z + x)(x + y)}}{{xyz}} \ge 8\]\[(1 + a^2 )(1 + b^2 )(1 + c^2 ) \ge \left[ {1 + \sqrt[3]{{\left( {abc} \right)^2 }}} \right]^3 \ge \left[ {1 + \sqrt[3]{{\left( 8 \right)^2 }}} \right]^3 = 125\] Generalization： Let $a_i > 0(i = 1,2, \cdots ,n),n \ge 3,m \in N,m \ge 1,$ such that \[\sum\limits_{i = 1}^n {\left( {\frac{1}{{1 + a_i }}} \right)} ^{\frac{1}{m}} = s \le 1,k > 0\] Prove that \[\prod\limits_{i = 1}^n {\left( {1 + a_i^k } \right)} \ge \left\{ {1 + \left[ {\left( {\frac{n}{s}} \right)^m - 1} \right]^k } \right\}^n \] someone please tell how to prove this generalization

Sayan wrote: As kunny pointed out, the given condition is equivalent to $abc \ge a+b+c+2$ Also note that \[1\ge \sum\frac1{1+a} \ge \frac3{\sqrt[3]{(1+a)(1+b)(1+c)}} \implies (1+a)(1+b)(1+c) \ge 27\]Therefore \[(a+b+c+3)^3 \ge 27(a+1)(b+1)(c+1) \ge 3^6 \implies a+b+c \ge 6\]This implies $abc \ge 8$ Hence by Holder (one can also use AM-GM) \[(1+a^2)(1+b^2)(1+c^2) \ge (1+(abc)^{\frac23})^3 \ge 125\] I used AM GM,on the whole thing(I mean I got 3 triplets and 1so I used AM GM on the triplets of terms)

Note that \[\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \le 1 \iff abc \ge a+b+c+2.\]And by CSE we have \[1\ge \sum \frac{1}{a+1} \ge \frac{9}{a+b+c+3} \implies a+b+c\ge 6 \implies abc\ge 8.\]By AM-GM we have \[\prod (1+a^2)=\prod \left(1+4\cdot \frac{a^2}{4}\right ) = \prod 5\sqrt[5]{\frac{a^8}{4^4}} =125. \]Equality occurs when $a=b=c=2$.

Titu lemma Errors: Instead of $\frac{9}{a+b+c}$, it is $\frac{9}{a+b+c+3}$

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