Question

For a > b > c 0 and a + b + c = 1,

Answer 1

Starting with what we know:
$a + b + c = 1$ is the same as:
$(a + b + c)^{2} = 1^{2}$
$a^{2} + b^{2} + c^{2} + 2(ab + bc + ac) = 1$

Now, let's use our inequality and bring it into our equation.
Since a > b and b >0, then a > 0
Thus, we can multiply both sides by a, b, or c since they are all positive values.

We can say: $ab > b^{2}$, since b > 0
Similarly, $ac > c^{2}$ and $bc > c^{2}$

Now, we've got values for ab, ac, and bc.
Using this information back into our original equation:
$a^{2} + b^{2} + c^{2} + 2(ab + bc + ac) = 1$

Since $ab > c^{2}$, then we can say:
$a^{2} + b^{2} + c^{2} + 2(ab + bc + ac) > a^{2} + b^{2} + c^{2} + 2(b^{2} + 2c^{2})$ and
$a^{2} + 3b^{2} + 5c^{2} < 1$ as required.