Show that a - b - c2a2 - b2 - c2-2bc - a - b - ca - b - c

#160; R.H.S. #10; #10; #10; #160; a+b+ca b c #10; #10; #10; #160; =a+b+ca ( #10;b+c )×a+( b+c )a+( b+c ) #160; #160; ...

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Question

Show that $\frac{{(a + b + c)}^{2}}{a^{2} + b^{2} + c^{2} - 2 b c} = \frac{a + b + c}{a - b + c}$

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a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

(a^{2} - 2 b c - b^{2} - c^{2}) \div (a - b - c) (a + b + c)

b^{2} c^{2} + c^{2} a^{2} + a^{2} b^{2} > a b c (a + b + c)

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} \geq a + b + c

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