Solution:
The given inequality may be written in the form

$$7 c^{2}-6(a+b) c-\left(a^{2}+b^{2}-6 a b\right)<0$$

Putting $x=7 c^{2}$, $y=-6(a+b) c$, $z=-\left(a^{2}+b^{2}-6 a b\right)$, we have to prove that $x+y+z<0$. Observe that $x, y, z$ are not all equal ($x>0, y<0$). Using the identity

$$x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$$

we infer that it is sufficient to prove $x^{3}+y^{3}+z^{3}-3 x y z<0$. Substituting the values of $x, y, z$, we see that this is equivalent to

$$343 c^{6}-216(a+b)^{3} c^{3}-\left(a^{2}+b^{2}-6 a b\right)^{3}-126 c^{3}(a+b)\left(a^{2}+b^{2}-6 a b\right)<0$$

Using $c^{3}=a^{3}+b^{3}$, this reduces to

$$343\left(a^{3}+b^{3}\right)^{2}-216(a+b)^{3}\left(a^{3}+b^{3}\right)-\left(a^{2}+b^{2}-6 a b\right)^{3}-126\left(a^{3}+b^{3}\right)(a+b)\left(a^{2}+b^{2}-6 a b\right)<0$$

This may be simplified (after some tedious calculations) to,

$$-a^{2} b^{2}\left(129 a^{2}-254 a b+129 b^{2}\right)<0$$

But $129 a^{2}-254 a b+129 b^{2}=129(a-b)^{2}+4 a b>0$. Hence the result follows.

1. We have

$$a^{3}=c^{3}-b^{3}=(c-b)\left(c^{2}+c b+b^{2}\right)$$

which is same as

$$\frac{a^{2}}{c-b}=\frac{c^{2}+c b+b^{2}}{a}$$

Similarly, we get

$$\frac{b^{2}}{c-a}=\frac{c^{2}+c a+a^{2}}{b}$$

We observe that

$$\frac{a^{2}}{c-b}+\frac{b^{2}}{c-a}=\frac{c\left(a^{2}+b^{2}\right)-a^{3}-b^{3}}{(c-a)(c-b)}=\frac{c\left(a^{2}+b^{2}-c^{2}\right)}{(c-a)(c-b)}$$

This shows that

$$\frac{a^{2}+b^{2}-c^{2}}{(c-a)(c-b)}=\frac{c^{2}+c b+b^{2}}{c a}+\frac{c^{2}+c a+a^{2}}{c b}$$

Thus it is sufficient to prove that

$$\frac{c^{2}+c b+b^{2}}{c a}+\frac{c^{2}+c a+a^{2}}{c b} \geq 6$$

However, we have $c^{2}+b^{2} \geq 2 c b$ and $c^{2}+a^{2} \geq 2 c a$. Hence

$$\frac{c^{2}+c b+b^{2}}{c a}+\frac{c^{2}+c a+a^{2}}{c b} \geq 3\left(\frac{b}{a}+\frac{a}{b}\right) \geq 3 \times 2=6$$

We have used AM-GM inequality.

2. Let us set $x=a / c$ and $y=b / c$. Then $x^{3}+y^{3}=1$ and the inequality to be proved is $x^{2}+y^{2}-1>6(1-x)(1-y)$. This reduces to

$$(x+y)^{2}+6(x+y)-8 x y-7>0$$

But

$$1=x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$$

which gives $x y=\left((x+y)^{3}-1\right) / 3(x+y)$. Substituting this in (1) and introducing $x+y=t$, the inequality takes the form

$$t^{2}+6 t-\frac{8}{3} \frac{\left(t^{3}-1\right)}{t}-7>0$$

This may be simplified to $-5 t^{3}+18 t^{2}-2 t+8>0$. Equivalently

$$-(5 t-8)(t-1)^{2}>0$$

Thus we need to prove that $5 t<8$. Observe that $(x+y)^{3}>x^{3}+y^{3}=1$, so that $t>1$. We also have

$$\left(\frac{x+y}{2}\right) \leq \frac{x^{3}+y^{3}}{2}=\frac{1}{2}$$

This shows that $t^{3} \leq 4$. Thus

$$\left(\frac{5 t}{8}\right)^{3} \leq \frac{125 \times 4}{512}=\frac{500}{512}<1$$

Hence $5 t<8$, which proves the given inequality.

3. We write $b^{3}=c^{3}-a^{3}$ and $a^{3}=c^{3}-b^{3}$ so that

$$c-a=\frac{b^{3}}{c^{2}-c a+a^{2}}, \quad c-b=\frac{a^{3}}{c^{2}-c b+b^{2}}$$

Thus the inequality reduces to

$$a^{2}+b^{2}-c^{2}>6 \frac{a^{3} b^{3}}{\left(c^{2}-c a+a^{2}\right)\left(c^{2}-c b+b^{2}\right)}$$

This simplifies (after some lengthy calculations) to

$$\begin{aligned} & -c^{6}-(a+b) c^{5}-a b c^{4}+\left(a^{3}+b^{3}\right) c^{3}+\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) c^{2} \\ & \quad+\left(a^{2} b+a b^{2}+a^{3}+b^{3}\right) a b c+\left(a^{4} b^{2}-6 a^{3} b^{3}+a^{2} b^{4}\right)>0 \end{aligned}$$

Substituting
$$c^{3}=a^{3}+b^{3}, \quad c^{4}=c\left(a^{3}+b^{3}\right), \quad c^{5}=c^{2}\left(a^{3}+b^{3}\right), \quad c^{6}=\left(a^{3}+b^{3}\right)^{2}$$
the inequality further reduces to
$$a^{2} b^{2}\left(a^{2}+b^{2}+c^{2}+a c+b c-6 a b\right)>0$$

Thus we need to prove that $a^{2}+b^{2}+c^{2}+a c+b c-6 a b>0$. Since $a^{2}+b^{2} \geq 2 a b$, it is enough to prove that $c^{2}+c(a+b)-4 a b>0$. Multiplying this by $c$ and using $a^{3}+b^{3}=c^{3}$, we need to prove that
$$a^{3}+b^{3}+c^{2} a+c^{2} b>4 a b c$$

Using AM-GM inequality to these 4 terms and using $c>a, c>b$ we get
$$a^{3}+b^{3}+c^{2} a+c^{2} b>4\left(a^{3} b^{3} c^{2} a c^{2} b\right)^{1 / 4}=4 a b c$$
which proves the inequality.