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

show that
a^(2) + 3b^(2) + 5c^(2) \ \textless \ 1

User Pwyg
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1 Answer

3 votes
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.
User Adambox
by
8.6k points
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