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Let a, b, c be real numbers such that a + b + c = 1. Find the minimum value of \(2a^2 + 3b^2 + 6c^2.\)

 Sep 1, 2019
 #1
avatar+4325 
+3

Try to use Cauchy-Schwarz Inequality because of the sum of squares. 

 Sep 2, 2019
 #2
avatar+23314 
+3

Let a, b, c be real numbers such that \(a + b + c = 1\).

Find the minimum value of \(2a^2 + 3b^2 + 6c^2\).

 

Schwarz Inequality:

\(\begin{array}{|rcll|} \hline \mathbf{(x_1^2+x_2^2+\ldots + x_n^2)(y_1^2+y_2^2+\ldots + y_n^2) } &\geq& \mathbf{ \left(x_1y_1+x_2y_2+\ldots x_ny_n \right)^2 } \\\\ \left( \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} \right)(2a^2+3b^2+6c^2 ) &\geq& \left( \dfrac{1}{\sqrt{2}} \sqrt{2}a+ \dfrac{1}{\sqrt{3}} \sqrt{3}b+ \dfrac{1}{\sqrt{6}} \sqrt{6}c \right)^2 \\\\ \left( \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} \right)(2a^2+3b^2+6c^2 ) &\geq& \left( a+b+c \right)^2 \\\\ \left( \underbrace{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}}_{=1} \right)(2a^2+3b^2+6c^2 ) &\geq& \left(\underbrace{ a+b+c}_{=1} \right)^2 \\ 2a^2+3b^2+6c^2 &\geq& 1^2 \\ \mathbf{2a^2+3b^2+6c^2} &\geq& \mathbf{1} \\ \hline \end{array}\)

 

The minimum value of \(2a^2 + 3b^2 + 6c^2 \)is \(\mathbf{1}\)

 

laugh

 Sep 2, 2019

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