Which of the following inequalities is true for any positive real numbers a, b and c ?
I. $$ab(a + b) + bc(b + c) + ca(c +a) \leq 6abc $$
II. $$\frac{a^{2} + b^{2} + c^{2}}{abc} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$
I) $$ab(a + b) + bc(b + c) + ca(c +a) \leq 6abc $$
$$\frac{ab(a+b)}{abc}+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}\le6$$
$$\frac{(a+b)}{c}+\frac{(b+c)}{a}+\frac{(c+a)}{b}\leq6$$
We shall rearrange the terms
$$\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{a}\le6$$
Here, we have six terms: three fractions and their reciprocals.
$$t+\frac{1}{t}\ge2$$ this is always true.
So three such equations add to $$\ge6$$
So, it is only sometimes valid.
II. $$\frac{a^{2} + b^{2} + c^{2}}{abc} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$
$$\frac{a^2+b^2+c^2}{abc}\le\frac{a\cdot b+b\cdot c+c\cdot a}{abc}$$
$$a^2+b^2+c^2\le a\cdot b+b\cdot c+c\cdot a$$
Consider (a,b,c) as (1,2,3)
Substituting them in the above equation,
$$1^2+2^2+3^2\le1\cdot2+2\cdot3+3\cdot1$$
$$14\le11$$ is not true, so this is also not true.
Hence, option D is correct.