khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that : $ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$ The best inequality of this type is the following: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation.

khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that : $ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$ Write the inequality as $ \sum \frac{a^2+b^2}{a^2+b^2+2} \ge \frac{3}{2},$ or $ \sum \frac{(a+b)^2}{(a+b)^2+\frac{2(a+b)^2}{a^2+b^2}} \ge \frac{3}{2}.$ Applying the Cauchy Schwarz Inequality, we have $ LHS \ge \frac{4(a+b+c)^2}{\sum (a+b)^2+2\sum \frac{(a+b)^2}{a^2+b^2}} =\frac{36}{\sum (a+b)^2+2\sum \frac{(a+b)^2}{a^2+b^2}}.$ It suffices to prove that $ \sum (a+b)^2+ 2\sum \frac{(a+b)^2}{a^2+b^2} \le 24.$ Because $ 12-\sum (a+b)^2 =\frac{4}{3} (a+b+c)^2 -\sum (a+b)^2 =-\frac{1}{3} \sum (a-b)^2,$ and $ 12-2\sum \frac{(a+b)^2}{a^2+b^2} =2\sum \frac{(a-b)^2}{a^2+b^2},$ this inequality is equivalent to $ \sum (a-b)^2 \left( \frac{6}{a^2+b^2} -1\right) \ge 0.$ Under the assumption that $ a \ge b \ge c,$ we see that this inequality is obviously true if $ a^2+b^2 \le 6.$ Let us consider now the case $ a^2+b^2 \ge 6,$ in this case we have $ \frac{1}{a^2+b^2+2} \le \frac{1}{8},$ and $ \frac{1}{a^2+c^2+2}+\frac{1}{b^2+c^2+2} \le \frac{1}{a^2+2}+\frac{1}{b^2+2} \le \frac{1}{8-b^2}+\frac{1}{b^2+2} \le \frac{1}{8}+\frac{1}{2},$ (because $ 0 \le b^2 \le \frac{9}{4}$) Hence $ \sum \frac{1}{a^2+b^2+2} \le \frac{1}{8}+\frac{1}{8}+\frac{1}{2} =\frac{3}{4}.$ Our proof is completed.

Vasc wrote: khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that : $ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$ The best inequality of this type is the following: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. Primitive mixing variables also can solve this one. Solution: WLOG $ a = max\{a,b,c\}$ \[ f(a,b,c) = \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)} \] \[ f(a,\frac {b + c}{2},\frac {b + c}{2}) = \frac 2{8 + 5(a^2 + \frac {(b + c)^2}{4})} + \frac 1{8 + 5\frac {(b + c)^2}{2}} \] \[{ \Rightarrow \ \ f(a,\frac {b + c}{2},\frac {b + c}{2}}) - f(a,b,c) = \] \[ (b - c)^2. \left[ \frac {\frac {5}{4}(16 + 10a^2 - 5b^2 - 5c^2 - 20bc)}{[8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)]} + \frac {\frac {5}{2}}{[8 + \frac {5(b + c)^2}{2}][8 + 5(b^2 + c^2]}\right] \] Note that : \[ [8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)] \ge [8 + \frac {5(b + c)^2}{2}][8 + 5(b^2 + c^2)]^2 \] \[{ \Rightarrow \ \ f(a,\frac {b + c}{2},\frac {b + c}{2}} - f(a,b,c)\ge \] \[ \ge (b - c)^2. \left[ \frac {\frac {5}{4}(16 + 10a^2 - 5b^2 - 5c^2 - 20bc) + \frac {5}{2}(8 + 5(b^2 + c^2)}{[8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)]} \right] \ge \] \[ \ge (b - c)^2. \left[ \frac {\frac {5}{4}(32 + 10a^2 - 5b^2 - 5c^2 - 20bc) + \frac {25(b^2 + c^2)}{2}}{[8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)]} \right] \ge 0 \] Therefore, we only need to prove the inequality in the case when b=c . Now it's a very simple.

Mixing Variables method is not enought for Vasc's problem... I think that. Because it only works for the case $ 0\le bc\le \frac{16}{20}$. But it can show you why we have the second equality. In fact we have: \[ f(a,b,c)=RHS-LHS\] And \[ f\left(x,\frac{3-x}{2},\frac{3-x}{2}\right)=\frac{5(x-1)^2(5x-13)^2}{6(25x^2-30x+77)(5x^2-30x+61)}\ge 0\]

khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that : $ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$ Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ .find constant k is best such that: $ \frac {1}{1 + k(a^{2} + b^{2})} + \frac {1}{1 + k(b^{2} + c^{2})} + \frac {1}{1 + k(c^{2} + a^{2})} \leq \frac {3}{1 + 2k}$ i think $ k<\frac{2}{3}$

Vasc wrote: khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that : $ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$ The best inequality of this type is the following: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. Since $ a+b+c = 3= const$. Fix $ a^2+b^2+c^2 = k = constant$ Consider function $ f=\frac{1}{8+5(k-a^2)}$ . $ f'''(a) = \frac{600a(5a^2+8+5k)}{(-8-5k+5a^2)^4} >0$ so apply n-1 EV, we only consider $ f(a,a,b)$. Now the problem is much simple and can be solved.

This Iran 2009 is an old result of Vasc http://www.mathlinks.ro/Forum/viewtopic.php?t=77182

khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that : $ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$ Here is my solution: $ Lemma$ :With $ x,y,z>0$,we have: $ (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2 \ge 2\sqrt{3(xy+yz+zx)}+4(x+y+z)$ $ Proof$ . Put $ m=\sqrt{b+c}+\sqrt{c+a}-\sqrt{a+b};n=\sqrt{c+a}+\sqrt{a+b}-\sqrt{b+c}$ $ p=\sqrt{a+b}+\sqrt{b+c}-\sqrt{c+a}$,then $ m,n,p>0$ This inequality become: $ mn+np+pm \ge \sqrt{3mnp(m+n+p)}$ Which is obvious true. Equality holds if and only if $ x=y=z$ -Now,com back the problem It is equivalent to: $ \frac{a^2+b^2}{2+a^2+b^2}+\frac{b^2+c^2}{2+b^2+c^2}+\frac{c^2+a^2}{2+c^2+a^2} \ge \frac{3}{2}$ By Cauchy-Schwarz inequality and Lemma,we have; $ LHS \ge \frac{(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2}{6+2(a^2+b^2+c^2)}$ $ \ge \frac{2(a^2+b^2+c^2)+ab+bc+ca}{3+a^2+b^2+c^2}$ (because:$ \sqrt{3(a^2b^2+b^2c^2+c^2a^2} \ge ab+bc+ca$) $ = \frac{\frac{3}{2}(3+a^2+b^2+c^2)}{3+a^2+b^2+c^2}=RHS$ (because $ a+b+c= 3$) Then,we have Q.E.d Equality holds if and only if $ a=b=c=1$

Your proof can be write into: $ LHS \geq\ \frac{(\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}})^{2}}{6+2(a^{2}+b^{2}+c^{2})}$ $ = \frac{2\sum\ a^2+2\sum\ \sqrt{(a^2+b^2)(a^2+c^2)}}{6+2(a^{2}+b^{2}+c^{2})}$ $ \geq\ \frac{2\sum\ a^2+2(\sum\ a^2+\sum\ ab)}{6+2(a^2+b^2+c^2)} = \frac{3}{2}$

@quykhtn-qa1: Click to reveal hidden text

nice solution,quykhtn-a1 Here is my solution Let $ f(a,b,c)=\frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}}$ => $ f(a,t,t)=\frac{2}{2+a^2+t^2}+\frac{1}{2+2t^2}$ with $ t=\frac{b+c}{2}$ We have $ f(a,t,t)-f(a,b,c)$ $ =(b^2+c^2-\frac{(b+c)^2}{2})(\frac{1}{(b^2+c^2+2)(2+\frac{(b+c)^2}{2})}$ $ -\frac{1}{(4+2a^2+b^2+c^2)(4+2a^2+\frac{(b+c)^2}{2})})\ge 0$ It's true because $ 2(b^2+c^2)\ge (b+c)^2$ and $ 4+2a^2+b^2+c^2\ge b^2+c^2+2,4+2a^2+\frac{(b+c)^2}{2}\ge 2+\frac{(b+c)^2}{2}$ And then,we prove that $ f(a,t,t)\le\frac{3}{4} \Leftrightarrow (a-1)^2(15a^2-78a+111)\ge 0$ (which is obvious true)(Q.E.D) Equality holds if and only if $ a=b=c=1$

Vasc wrote: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. This inequality is true for all reals $ a,$ $ b$ and $ c$ such that $ a + b + c = 3.$ Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2$ and $ abc = w^3.$ Since, $ \prod_{cyc}(a^2 + b^2) = - w^6 + A(u,v^2)w^3 + B(u,v^2)$ we see that the original inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is convex function. Hence, $ f$ gets a maximal value, when $ w^3$ gets an extremal value, which happens when two numbers from $ \{a,b,c\}$ are equal. Thus, for the poof enough to check only one case: $ b = c = \frac {3 - a}{2},$ which gives $ (a - 1)^2(5a - 13)^2\geq0.$ Done!

arqady wrote: Vasc wrote: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. This inequality is true for all reals $ a,$ $ b$ and $ c$ such that $ a + b + c = 3.$ Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2$ and $ abc = w^3.$ Since, $ \prod_{cyc}(a^2 + b^2) = - w^6 + A(u,v^2)w^3 + B(u,v^2)$ we see that the original inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is convex function. Hence, $ f$ gets a maximal value, when $ w^3$ gets an extremal value, which happens when two numbers from $ \{a,b,c\}$ are equal. Thus, for the poof enough to check only one case: $ b = c = \frac {3 - a}{2},$ which gives $ (a - 1)^2(5a - 13)^2\geq0.$ Done! Maybe you mean $ f(u,v^2,w^3)\leq 0$ and $ f$ is convex?

Karamata wrote: Maybe you mean $ f(u,v^2,w^3)\leq 0$ and $ f$ is convex? Otherwise, it doesn't follow that $ f$ will attain its maximum only if $ w^3$ is extremal $ f$ is convex function of $ w^3.$

This is also my proof, Arqady. We can generalize this inequality to $ n$ variables as follows: Let $ a_1,a_2,...a_n$ be real numbers such that $ a_1 + a_2 + ... + a_n = n$. If $ r\ge \frac {n^2 - 1}{n^2 - n - 1}$, then $ \sum \frac 1{r + a_2^2 + ... + a_n^2} \le \frac n{r + n - 1}$, with equality for $ a_1 = ... = a_n = 1$, and for $ a_1 = ... = a_{n - 1} = \frac 1{n^2 - n - 1}$ and $ a_n = n - \frac {n - 1}{n^2 - n - 1}$.

Another solution

Attachments:

$ a,b,c > 0 $ and $a+b+c=3$ $\frac{1}{3+a^{2}+b^{2}}+\frac{1}{3+b^{2}+c^{2}}+\frac{1}{3+c^{2}+a^{2}}\leq\frac{3}{5}$ the ineq becomes to $\sum{\frac{a^2+b^2}{3+a^2+b^2}} \geq \frac{6}{5}$ and by C-S also have $\left(\sum{\frac{a^2+b^2}{3+a^2+b^2}}\right)(\sum{3+a^2+b^2}) \geq (\sum{\sqrt{a^2+b^2}})^2$ That means we have to prove $(\sum{\sqrt{a^2+b^2}})^2 \geq \frac{6}{5} (\sum{(3+a^2+b^2)}) $ $\sum{(a^2+b^2)}+2\sum{\sqrt{(a^2+b^2)(a^2+c^2)}} \geq \frac{54}{5} + \frac{12}{5}\sum{a^2}$ $8\sum{a^2} + 10\sum{ab} \geq 54$ $5(a+b+c)^2 + 3\sum{a^2} \geq 54$ and finish the prove.

arqady wrote: Vasc wrote: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. ....Hence, $ f$ gets a maximal value, when $ w^3$ gets an extremal value, which happens when two numbers from $ \{a,b,c\}$ are equal. Dear Arqady, according to the $uvw$ method, don't you need to check also the case $a=0$? Thank you, best regards, Arrigo-Sacchi

Arrigo-Sacchi wrote: arqady wrote: Vasc wrote: If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. ....Hence, $ f$ gets a maximal value, when $ w^3$ gets an extremal value, which happens when two numbers from $ \{a,b,c\}$ are equal. Dear Arqady, according to the $uvw$ method, don't you need to check also the case $a=0$? Thank you, best regards, Arrigo-Sacchi No! Because this inequality is true for all reals $a$, $b$ and $c$ such that $a+b+c=3$. See my full post: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1790287#p1790287

Vasc wrote: The best inequality of this type is the following: If $ a,b,c$ are real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)} + \frac 1{8 + 5(a^2 + b^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. $\frac{1}{6}-\sum{\frac{1}{8+5b^2+5c^2}}$ $=\frac{125}{27}\prod{\frac{(b-c)^2}{8+5b^2+5c^2}}+\frac{5}{54}\sum{\frac{(1-5a)^2(b-c)^2}{(8+5c^2+5a^2)(8+5a^2+5b^2)}}\geq0.$ http://www.artofproblemsolving.com/Forum/viewtopic.php?t=359320

Suppose that $ a$,$ b$,$ c$ be three nonnegative real numbers such that $ a+b+c=2$ . Prove that : $$ \frac{1}{1+a^{2}+b^{2}}+\frac{1}{1+b^{2}+c^{2}}+\frac{1}{1+c^{2}+a^{2}} \geq \frac{4}{3}$$Where?

Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ ab+bc+ca=1$ . Prove that$$\frac{a}{2+b^{2}+c^{2}}+\frac{b}{2+c^{2}+a^{2}}+ \frac{c}{2+a^{2}+b^{2}} \geq \frac{3\sqrt{3}}{8}$$(Vietnam)

Ji Chen wrote: Vasc wrote: The best inequality of this type is the following: If $ a,b,c$ are real numbers such that $ a + b + c = 3$, then $ \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)} + \frac 1{8 + 5(a^2 + b^2)}\le \frac 1{6}$, with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation. $\frac{1}{6}-\sum{\frac{1}{8+5b^2+5c^2}}$ $=\frac{125}{27}\prod{\frac{(b-c)^2}{8+5b^2+5c^2}}+\frac{5}{54}\sum{\frac{(1-5a)^2(b-c)^2}{(8+5c^2+5a^2)(8+5a^2+5b^2)}}\geq0.$ http://www.artofproblemsolving.com/Forum/viewtopic.php?t=359320 szl,note jichen this post..

khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that : $ \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4}$ Proof : By C-S and AM-GM, $$\sum \frac{1}{2+a^2+b^2}=\frac{3}{2}-\frac{1}{2}\sum \frac{a^2+b^2}{2+a^2+b^2}\leq \frac{3}{2}-\sum \frac{(\sum \sqrt{a^2+b^2})^2}{4\sum a^2+12}\leq \frac{3}{2}-\sum \frac{2\sum a^2+\sum (a+b)(a+c)}{4\sum a^2+12}$$$$=\frac{3}{2}-\sum \frac{3\sum a^2+3\sum ab}{4\sum a^2+12}\leq \frac{3}{2}-\sum \frac{3\sum a^2+9}{4\sum a^2+12}=\frac{3}{4}.$$

Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that$$\frac{1}{a^2-b +4}+\frac{1}{b^2-c +4}+\frac{1}{c^2-a +4} \ge \frac{3}{4}$$p/6487921617 Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that$$\frac {1}{a^2+b^2+3} + \frac {1}{b^2+c^2+3} +\frac{1 } {c^2+a^2+3}\leq \frac {3}{5}$$Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ abc(a+b+c)=3$ . Prove that$$\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+ \frac{1}{c^2+a^2+1}\leq 1$$

sqing wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that$$\frac{1}{a^2-b +4}+\frac{1}{b^2-c +4}+\frac{1}{c^2-a +4} \ge \frac{3}{4}$$ here

Attachments:

Let $a,b,c>0$ and $ab+bc+ca=1.$ Prove that $$\frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\leq\frac{9}{5}$$Let $a,b,c>0$ and $a^2+b^2+c^2=1.$ Prove that $$\frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\geq\frac{9}{5}$$Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$\frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\leq\frac{27}{11}$$Let $a,b,c>0$ and $abc=1.$ Prove that $$\frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\leq1$$Let $a,b,c>0$ and $a+b+c=2.$Prove that $$\frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\geq\frac{4}{3}$$Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that$$ \frac{a^2}{b^2-b+1}+\frac{b^2}{c^2-c+1}+\frac{c^2}{a^2-a+1}\geq 3$$

Attachments:

Redacted

khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that : $ \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4}$ Because$:$ \begin{align*} &571536\, \prod \left( {a}^{2}+{b}^{2}+2 \right) \cdot (\text{RHS}-\text{LHS}) \\&=3\sum\limits_{cyc} \, \left( 1524\,{a}^{2}+32\,ac+1780\,{b}^{2}+63\,{c}^{2} \right) \left( a+b-2\,c \right) ^{4}\\&\,\,\,+81\, \sum\limits_{cyc} \left( 1859\,{a}^{2}{b}^{2}+503\,ab {c}^{2}+3619\,a{c}^{3}+96\,{b}^{3}c+3331\,b{c}^{3} \right) \left( a-b \right) ^{2} \\& \geq 0\end{align*}

khashi70 wrote: Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that : $$ \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4}$$ Solution of Zhangyanzong: $$\sum_{cyc}\frac{a^2+b^2}{2+a^2+b^2}\geq \frac{(\sum_{cyc}\sqrt{a^2+b^2})^2}{2\sum_{cyc} a^2+6}=\frac{2\sum_{cyc} a^2+2\sum_{cyc}\sqrt{(a^2+b^2)(a^2+c^2)}}{2\sum_{cyc} a^2+6}\geq\frac{2\sum_{cyc} a^2+2\sum_{cyc}(a^2+bc)}{2\sum_{cyc} a^2+6} =\frac{3}{2} $$$$\sum_{cyc}\frac{1}{2+a^2+b^2}\leq \frac{3}{4}$$https://artofproblemsolving.com/community/c6h375652p19876954 Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that :

$$\frac{1}{(b+c)^2+a^2}+\frac{1}{(c+a)^2+b^2}+\frac{1}{(a+b)^2+c^2}\leq \frac{3}{5}$$S

If $a, b$ and $c$ are positive then$$\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$$

Let $ a,b,c>0$ and $\frac{1}{a^2+b^2+2}+ \frac{1}{b^2+c^2+2}+\frac{1}{c^2+a^2+2}=1 .$ Prove that $$ a+b+c\leq \frac{3}{\sqrt 2}$$Let $ a,b>0$ and $\frac{1}{a^2+b^2+2}+\frac{1}{a^2+2}+\frac{1}{b^2+2} =1 .$ Prove that $$ a+b\leq \sqrt{\sqrt{17} -1}$$$$ ab\leq \frac{\sqrt{17}-1}{4}$$here

Let $ a,b,c>0$ and $\frac{1}{a^2+b^2+1}+ \frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\geq 1 .$ Prove that $$ b^2+3 \ge 2b(a+c)$$$$ab + bc + ca \leq 3$$here

Let $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3 .$ Prove that $$ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)}\geq \frac{642}{4697}$$with equality for $ (a,b,c)=\left(\frac 3{2},\frac 3{2},0\right).$ https://artofproblemsolving.com/community/c4h3108842p28127400

Hope this is correct... We replace $b^2+c^2=2x,a^2+c^2=2y,a^2+b^2=2z$ which yields \[3xyz +\sum{xy}\overset{?}{\geq} \sum{x}+3\]with condition $\sqrt{y+z-x}+\sqrt{x+z-y}+\sqrt{x+y-z}=3$. This one is equavilent to (with different condition) \[\sum{\sqrt{x}}=3\overset{?}{\implies} 3\Pi{(y+z)}+2\sum{x^2}+6\sum{yz}\overset{?}{\geq} 8\sum{x}+24\]Let $x+y+z=3u,xy+yz+zx=3v^2,xyz=w^3$. \[3(9uv^2-w^3)+2(9u^2-6v^2)+18v^2\overset{?}{\geq} 24u+24 \iff 9uv^2+6u^2+2v^2-8u-8\overset{?}{\geq} w^3\]We have $u\geq 1$ and $u+v^2\geq 2$. If $u\geq 2$, then \[9uv^2+6u^2+2v^2-8u-8\geq w^3+12u+2v^2-8u-8\geq w^3+4u-8\geq w^3\]Suppose that $u<2$. \[9uv^2+6u^2+2v^2-8u-8-w^3\geq 8uv^2+6u^2+2v^2-8u-8\overset{?}{\geq} 0\]\[4uv^2+3u^2+v^2\overset{?}{\geq} 4u+4 \ \ \text{or} \ \ 4uv^2+3u^2+v^2\geq (4u+1)(2-u)\overset{?}{\geq} -3u^2+4u+4\]\[-4u^2-u+8u+2\overset{?}{\geq} -3u^2+4u+4\iff u^2-3u+2=(u-1)(u-2)\overset{?}{\leq} 0\]Which is true by our assumption as desired.$\blacksquare$ Thanks below!

bin_sherlo wrote: \[\sum{\frac{1}{b^2+c^2+1+1}}\overset{CS}{\leq} \sum{\frac{1+1+a+1}{(a+b+c+1)^2}}=\frac{\sum{a}+9}{16}=\frac{3}{4}\]As desired.$\blacksquare$ I think the C-S is $$(b^2+c^2+1+1)(1+1+a^2+1) \geq (a+b+c+1)^2,$$And since $a^2+b^2+c^2 \geq 3,$ this isn't strong enough.