Show $a^2+b^2+c^2=1$ $\implies$ $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $

Question

If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that

$ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $

( don't apply Schur's inequality )? Thanks.

Answer 1

Let $x=a^2$, $y=b^2$, $z=c^2$ and $f(x,y,z)=\frac{yz}{1+x}+\frac{xz}{1+y}+\frac{xy}{1+z}$ and g(x,y,z)=x+y+z. Consider the restriction $\{(x,y,z) \in \mathbb{R}^3;g(x,y,z)=1\}$. Applying the method of Lagrange multipliers, we have

$$\frac{y}{1+z}+\frac{z}{1+y}-\frac{yz}{(1+x)^2} = \lambda$$ $$\frac{z}{1+x}+\frac{x}{1+z}-\frac{xz}{(1+y)^2} = \lambda$$ $$\frac{x}{1+y}+\frac{y}{1+x}-\frac{xy}{(1+z)^2} = \lambda$$ $$x+y+z=1$$

Subtracting the first two equations, we obtain

$$\frac{y-x}{1+z}+\frac{z}{1+y}(1-\frac{x}{1+y})-\frac{z}{1+x}(1+\frac{y}{1+x})=0 \Rightarrow$$ $$(y-x) \left( \frac{1}{1+z}+\frac{z(x+y+1)(x+y+2)}{(1+x)^2(1+y)^2} \right )=0 $$

The second factor is $\ge 0$, so $x=y$. Analogously, $y=z$. Thus, we conclude that $x=y=z=1/3$. How $f(1/3,1/3,1/3)=1/4$ and $f$ have a maximum in this restriction, we have the desired inequality.

Answer 2

$x=3a^2,y=3b^2,z=3c^2 \implies x+y+z=3 ,\dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} =\dfrac {yz}{3(3+x)} +\dfrac {xz}{3(3+y)}+\dfrac {xy}{3(3+z)} \le \dfrac 14 \iff 4\sum_{cyc}xy(3+x)(3+y) \le 3(3+x)(3+y)(3+z) --<1>$

Now we use UVW method:

$3u=x+y+z=3,3v^2=xy+yz+xz,w^3=xyz \implies u=1 ,u \ge v \ge w$

$\sum_{cyc}xy(3+x)(3+y)=\sum_{cyc}xy(9+3(x+y)+xy)=\sum_{cyc}xy(9+3(3-z)+xy)=\sum_{cyc}[xy(18+xy)-3xyz]=\sum_{cyc}[18xy+(xy)^2]-9xyz=18*3v^2+[(3v^2)^2-2xyz(x+y+z)]-9w^3=54v^2+9v^4-15w^3$

$3(3+x)(3+y)(3+z)=81+27(x+y+z)+9(xy+yz+xz)+3xyz=81*2+27v^2+3w^3$

$<1> \iff 21v^2+4v^4 \le18+7w^3$

$w^3\ge 3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}= 3v^2-2-2\sqrt{(1-v^2)^3}$

case 1:

if $ 3v^2-2-2\sqrt{(1-v^2)^3} \le 0 \implies v^2 \le \dfrac{3}{4} \implies 21v^2+4v^4 \le 18 $ the inequality is always true.

case 2:

when $3v^2-2-2\sqrt{(1-v^2)} \ge 0 \implies v^2 \ge \dfrac{3}{4} , 18+7w^3 \ge 4+21v^2-14\sqrt{(1-v^2)^3} $

$<1> \iff 2-2v^4-7(1-v^2)\sqrt{1-v^2} \ge 0 \iff (1-v^2)(2+2v^2-7\sqrt{1-v^2}) \ge 0 $

$1-v^2 \ge 0$ when $v^2=1$ get "="

$2+2v^2-7\sqrt{1-v^2} \ge 0 \iff 4v^4 + 57v^2-45 \ge 0 \iff (4v^2-3)(v^2+15)\ge 0 \iff v^2 \ge \dfrac{3}{4}$

it is also true and when $v^2=\dfrac{3}{4}$ get "="

so there are two solutions to get "=":

$v^2=1 \implies w^3=1 \implies x=y=z$

$v^2=\dfrac{3}{4} \implies w^3=0, x=0,y=z=\dfrac{3}{2}$ or cycle

Answer 3

By C-S $$\sum_{cyc}\frac{b^2c^2}{1+a^2}=\sum_{cyc}\frac{b^2c^2}{a^2+b^2+a^2+c^2}\leq$$ $$\leq\sum_{cyc}\frac{b^2c^2}{4}\left(\frac{1}{a^2+b^2}+\frac{1}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{b^2c^2}{a^2+c^2}\right)=$$ $$=\frac{1}{4}\sum_{cyc}\left(\frac{b^2c^2}{a^2+b^2}+\frac{a^2c^2}{a^2+b^2}\right)=\frac{1}{4}.$$ Done!