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# help needed, thank you in advanced

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Let a, b, c be real numbers such that min(a + 24b + 38c, a + 16b + 42c) ≥ 2021. Prove that a^2 + b^2 + c^2 > 2021.

I know that we have to break it into two cases, one when a + 24b + 38c is the minimum, and the other when  a + 16b + 42c is the minimum, but I am not quite sure how to proceed from there.

Jan 16, 2021

#1
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Can't say that I fully understand this question, but here's some algebra that might help.

Consider

$$\displaystyle (a-1)^{2}+(b-24)^{2}+(c-38)^{2} ,$$

(geater than or equal to zero),  for all real values of a, b and c.

Expanding each bracket and rearranging,

$$\displaystyle a^{2}+b^{2}+c^{2}-2(a+24b+38c)+1^{2}+24^{2}+38^{2} \geq0, \\ \text{so} \\ a^{2}+b^{2}+c^{2}+2021 \geq2(a+24b+38c), \\ \text{and since } \\ a+24b+38c \geq 2021, \\a^{2}+b^{2}+c^{2} \geq 2(2021)-2021=2021.$$

The other part, a + 16b + 42c, leads to exactly the same result, but how that's to be presented to answer the question, I don't know.

Jan 17, 2021
#2
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Thanks Tiggsy.

I have been playing, unsuccessfully with this one.

Never would have thought of that.   :)

Melody  Jan 17, 2021