I can't believe I had to do this again. Claim: $\tfrac{1}{1 + c^2} \geq 1 - \tfrac c2$ for $c \geq 0$. Proof: Just expand, and get $2 \geq (2-c)(1 + c^2) \iff c(c-1)^2 \geq 0$. Equality when $c = 0, 1$. Hence, $\tfrac{1}{1 + x^4} \geq \tfrac{2 - x^2}{2}$ with equality at $x = 1$ for $x > 0$. Write\[\sum_{cyc} \frac{bc}{1 + a^4} \geq \sum \left(bc - \frac{a^2bc}{2} \right) = 3 - \frac12abc(a + b + c)\]so it suffices to show\[abc(a + b + c) \leq 3 \iff 3abc(a + b + c) \leq (ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c)\]which reduces to\[abc(a + b + c) \leq a^2b^2 + b^2c^2 + c^2a^2\]which is simply just the known inequality $xy + yz + zx \leq x^2 + y^2 + z^2$ for $x = ab, y = bc, z = ca$, so we are done.

not needed ..

$$bc=x,ab=z,ac=y$$$\frac{bc}{1+a^4} = \frac{x^3}{x^2+y^2z^2}= x - \frac{xy^2z^2}{x^2+y^2z^2}$ $$\sum \frac{xy^2z^2}{x^2+y^2z^2} \leq \sum \frac{xy^2z^2}{2xyz} \leq \frac{3}{2}$$$\blacksquare$

Let $a, b,$ and $c$ be positive real numbers such that $ab + bc + ca = 3$. Show that$$\dfrac{bc}{1 + a^2} + \dfrac{ca}{1 + b^2} + \dfrac{ab}{1 + c^2} \geq \dfrac{3}{2}$$

sqing wrote: Let $a, b,$ and $c$ be positive real numbers such that $ab + bc + ca = 3$. Show that$$\dfrac{bc}{1 + a^2} + \dfrac{ca}{1 + b^2} + \dfrac{ab}{1 + c^2} \geq \dfrac{3}{2}$$

jj_ca888 wrote: I can't believe I had to do this again. Claim: $\tfrac{1}{1 + c^2} \geq 1 - \tfrac c2$ for $c \geq 0$. Proof: Just expand, and get $2 \geq (2-c)(1 + c^2) \iff c(c-1)^2 \geq 0$. Equality when $c = 0, 1$. Hence, $\tfrac{1}{1 + x^4} \geq \tfrac{2 - x^2}{2}$ with equality at $x = 1$ for $x > 0$. Write\[\sum_{cyc} \frac{bc}{1 + a^4} \geq \sum \left(bc - \frac{a^2bc}{2} \right) = 3 - \frac12abc(a + b + c)\]so it suffices to show\[abc(a + b + c) \leq 3 \iff 3abc(a + b + c) \leq (ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c)\]which reduces to\[abc(a + b + c) \leq a^2b^2 + b^2c^2 + c^2a^2\]which is simply just the known inequality $xy + yz + zx \leq x^2 + y^2 + z^2$ for $x = ab, y = bc, z = ca$, so we are done. Solution indeed. Bravo

InternetPerson10 wrote: Let $a, b,$ and $c$ be positive real numbers such that $ab + bc + ca = 3$. Show that \[ \dfrac{bc}{1 + a^4} + \dfrac{ca}{1 + b^4} + \dfrac{ab}{1 + c^4} \geq \dfrac{3}{2}. \]

Let $x = bc, y = ca, z = ab$, then $x + y + z = 3$. Then $\sum_{cyc} \frac{bc}{1+a^4} = \sum_{cyc} \frac{x}{1+(yz/x)^2} = \sum_{cyc} \frac{x^3}{x^2 + y^2z^2} = \sum_{cyc} x - \frac{xy^2z^2}{x^2 + y^2z^2} = 3 - \sum_{cyc} \frac{xy^2z^2}{x^2 + y^2z^2} \ge 3 - \sum_{cyc} \frac{xy^2z^2}{2xyz} = 3 - \frac{1}{2}(xy + yz + zx) \ge 3 - \frac{1}{2} \cdot \frac{(x+y+z)^2}{3} = \frac{3}{2}.$

Here is a brute forced solution XD (took less time than i thought) Using Cauchy we have that: $$\frac{bc}{1+a^4}+\frac{ac}{1+b^4}+\frac{ab}{1+c^4} \ge \frac{9}{3+abc(a^3+b^3+c^3)}$$So now we need to prove that $$\frac{9}{3+abc(a^3+b^3+c^3)} \ge \frac{3}{2} \; \; \text{which is equivalent to} \; \; 3 \ge abc(a^3+b^3+c^3)$$So now we use Lagrange multipliers, define $g(a,b,c)=ab+bc+ac-3$ and $f(a,b,c)=a^3+b^3+c^3$ so now, for a constant $\lambda \in \mathbb R$ we define $\mathcal L(a,b,c,\lambda)=f(a,b,c)+\lambda g(a,b,c)=a^3+b^3+c^3+\lambda(ab+bc+ca-3)$ so now we need the gradient to be $0$ which is equivalent to find $a,b,c,\lambda$ so that $ab+bc+ac=3$ and $$3a^2=-\lambda(b+c)$$$$3b^2=-\lambda(a+c)$$$$3c^2=-\lambda(a+b)$$If $a \ne b \ne c$ then by substtracting one of those equations with the other we end up with $3a+3c=3b+3c=3a+3b=\lambda$ which means $a=b=c$ which is a contradiction so at least from the numbers $a,b,c$ one is equal to the other one, lets say $a=b$ WLOG so now if $a \ne c$ then by substracting 2 equations we get $3a+3c=\lambda$ but also in one of the equations we get $\lambda=\frac{-3c^2}{2a}$ but negative cant be equal to positive so we get a contradiction, meaning that $a=b=c$ so $a=b=c=1$ so the minimun of $f$ down condition $g$ is $f(1,1,1)=3$ so $3 \ge a^3+b^3+c^3$ and now by Am-Gm we have $3=ab+bc+ac \ge 3 (abc)^{\frac{2}{3}}$ which means $1 \ge abc$ so $$3 \ge a^3+b^3+c^3 \ge abc(a^3+b^3+c^3)$$Thus we are done

MathLuis wrote:

For Lagrange Multipliers, you have to show that a minimum of $f$ exists.

Here is my solution:

InternetPerson10 wrote: Let $a, b,$ and $c$ be positive real numbers such that $ab + bc + ca = 3$. Show that \[ \dfrac{bc}{1 + a^4} + \dfrac{ca}{1 + b^4} + \dfrac{ab}{1 + c^4} \geq \dfrac{3}{2}. \] We need to prove that $\sum \frac{bc}{a^4+1}-bc= -\sum \frac{bca^4}{a^4+1}\geq -\frac{3}{2}\Leftrightarrow \sum \frac{bca^4}{a^4+1}\leq \frac{3}{2}$ with is true because: $\sum \frac{bca^4}{a^4+1}\leqslant \sum \frac{bca^4}{2a^2}= abc(a=b+c)\leqslant \frac{(ab+bc+ca)^2}{3}=3$