Indian National Mathematical Olympiad

Solution:
We begin with the observation that
$$x^{2}+x y+y^{2}=\frac{3}{4}(x+y)^{2}+\frac{1}{4}(x-y)^{2} \geq \frac{3}{4}(x+y)^{2}$$
and similar bounds for $y^{2}+y z+z^{2}, z^{2}+z x+x^{2}$. Thus
$$3\left(x^{2}+x y+y^{2}\right)\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right) \geq \frac{81}{64}(x+y)^{2}(y+z)^{2}(z+x)^{2}$$
Thus it is sufficient to prove that
$$(x+y+z)(x y+y z+z x) \leq \frac{9}{8}(x+y)(y+z)(z+x)$$
Equivalently, we need to prove that
$$8(x+y+z)(x y+y z+z x) \leq 9(x+y)(y+z)(z+x)$$
However, we note that
$$(x+y)(y+z)(z+x)=(x+y+z)(y z+z x+x y)-x y z$$
Thus the required inequality takes the form
$$(x+y)(y+z)(z+x) \geq 8 x y z$$
This follows from AM-GM inequalities;
$$x+y \geq 2 \sqrt{x y}, \quad y+z \geq 2 \sqrt{y z}, \quad z+x \geq 2 \sqrt{z x}$$

Solution:
Let us introduce $x+y=c, y+z=a$ and $z+x=b$. Then $a, b, c$ are the sides of a triangle. If $s=(a+b+c) / 2$, then it is easy to calculate $x=s-a, y=s-b, z=s-c$ and $x+y+z=s$. We also observe that
$x^{2}+x y+y^{2}=(x+y)^{2}-x y=c^{2}-\frac{1}{4}(c+a-b)(c+b-a)=\frac{3}{4} c^{2}+\frac{1}{4}(a-b)^{2} \geq \frac{3}{4} c^{2}$.
Moreover, $x y+y z+z x=(s-a)(s-b)+(s-b)(s-c)+(s-c)(s-a)$. Thus it is sufficient to prove that
$$s \sum(s-a)(s-b) \leq \frac{9}{8} a b c$$
But, $\sum(s-a)(s-b)=r(4 R+r)$, where $r, R$ are respectively the in-radius, the circum-radius of the triangle whose sides are $a, b, c$, and $a b c=4 R r s$. Thus the inequality reduces to
$$r(4 R+r) \leq \frac{9}{2} R r$$
This is simply $2 r \leq R$. This follows from $I O^{2}=R(R-2 r)$, where $I$ is the incentre and $O$ the circumcentre.

Solution:
If we set $x=\lambda a, y=\lambda b, z=\lambda c$, then the inequality changes to
$$(a+b+c)^{2}(a b+b c+c a)^{2} \leq 3\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)$$
This shows that we may assume $x+y+z=1$. Let $\alpha=x y+y z+z x$. We see that
$$\begin{aligned} x^{2}+x y+y^{2} & =(x+y)^{2}-x y \\ & =(x+y)(1-z)-x y \\ & =x+y-\alpha=1-z-\alpha \end{aligned}$$
Thus
$$\begin{aligned} \prod\left(x^{2}+x y+y^{2}\right) & =(1-\alpha-z)(1-\alpha-x)(1-\alpha-y) \\ & =(1-\alpha)^{3}-(1-\alpha)^{2}+(1-\alpha) \alpha-x y z \\ & =\alpha^{2}-\alpha^{3}-x y z \end{aligned}$$
Thus we need to prove that $\alpha^{2} \leq 3\left(\alpha^{2}-\alpha^{3}-x y z\right)$. This reduces to
$$3 x y z \leq \alpha^{2}(2-3 \alpha)$$
However
$$3 \alpha=3(x y+y z+z x) \leq(x+y+z)^{2}=1$$
so that $2-3 \alpha \geq 1$. Thus it suffices to prove that $3 x y z \leq \alpha^{2}$. But
$$\begin{aligned} \alpha^{2}-3 x y z & =(x y+y z+z x)^{2}-3 x y z(x+y+z) \\ & =\sum_{\text {cyclic }} x^{2} y^{2}-x y z(x+y+z) \\ & =\frac{1}{2} \sum_{\text {cyclic }}(x y-y z)^{2} \geq 0 \end{aligned}$$