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Yuki888 [10]
3 years ago
7

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

Mathematics
1 answer:
DiKsa [7]3 years ago
3 0
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.
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