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# Algebra - Guide me, please!

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Let a, b and c be positive real numbers such that a + b + c = 1.

Find the minimum value of a^2 + 2(b^2) + c^2

I think this uses something known as the Cuachy-Schwartz inequality. However, i'm not sure what these means! Could you please guide me through this problem?

Thank you!

Dec 23, 2020

#1
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By Cauchy-Schwarz, to get the minimum value, let a = b = c = 1/3.  So the minimum value is (1/3)^2 + 2*(1/3)^2 + (1/3)^2 = 4/9.

Dec 24, 2020
#2
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These were my first steps that I tried applying to the problem, too. However, even though it seems intuitively true, it is not the correct answer. Perhaps values other than 1/3 would result in lower numbers?

That's where I'm stuck. I can't find anything else that could be lower.

Guest Dec 24, 2020
#3
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OP here.

I actually managed to get this problem! Here, a = c = 2/5 and b = 1/5.

the expression goes to 2/5. this was scary! thank you!

Dec 24, 2020
#4
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Treating the problem as a calculus max/min problem gets the result easily enough, but in keeping with the suggestion in the question here's the Cauchy-Schwarz solution.

For n = 3, Cauchy-Schwarz says that

$$\displaystyle a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \leq(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{1/2}(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})^{1/2},$$

so,

$$\displaystyle 1 =a+b+c=a.1+(\sqrt{2}b)(1/\sqrt{2})+c.1\leq(a^{2}+2b^{2}+c^{2})^{1/2}(1+\frac{1}{2}+1)^{1/2},\\ 1\leq(a^{2}+2b^{2}+c^{2})^{1/2}(5/2)^{1/2},$$

so squaring and cross-multiplying,

$$\displaystyle a^{2}+2b^{2}+c^{2}\geq2/5.$$

Dec 24, 2020