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# Inequalities

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If a,b,c are the sides of a triangle then how do I prove that $$(\dfrac {ab+bc+ca}{a^2+b^2+c^2})\geq1/2$$?

Potato  Nov 23, 2018
edited by Potato  Nov 23, 2018
#1
+20578
+8

If a,b,c are the sides of a triangle then how do I prove that

$$\large{\dfrac {ab+bc+ca}{a^2+b^2+c^2} \geq \dfrac12 }$$ ?

$$\begin{array}{|lrcll|} \hline 1. & a & \leq & b+c \quad | \quad \cdot a \\ & a^2 & \leq & (b+c)a \\ & \mathbf{a^2} & \mathbf{ \leq } & \mathbf{ ab+ca } \qquad (1) \\\\ 2. & b & \leq & a+c \quad | \quad \cdot b \\ & b^2 & \leq & (a+c)b \\ & \mathbf{b^2} & \mathbf{ \leq } & \mathbf{ab+bc} \qquad (2) \\\\ 3. & c & \leq & a+b \quad | \quad \cdot c \\ & c^2 & \leq & (a+b)c \\ & \mathbf{c^2} & \mathbf{ \leq } & \mathbf{ca+bc} \qquad (3) \\ \\ \hline \\ (1)+(2)+(3): & a^2+b^2+c^2 & \leq & ab+ca+ab+bc+ ca+bc \\ & a^2+b^2+c^2 & \leq & 2\cdot (ab+bc+ca) \quad | \quad : (a^2+b^2+c^2) \\ & 1 & \leq & 2\cdot \dfrac{ab+bc+ca} {a^2+b^2+c^2} \quad | \quad :2 \\ & \dfrac12 & \leq & \dfrac{ab+bc+ca} {a^2+b^2+c^2} \\ &\mathbf{ \dfrac{ab+bc+ca} {a^2+b^2+c^2}} & \mathbf{ \geq } & \mathbf{ \dfrac12 } \\ \hline \end{array}$$

heureka  Nov 23, 2018
#2
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Thank you😊

Potato  Nov 23, 2018
#3
+92429
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Nice, Heureka....!!!!

CPhill  Nov 23, 2018