Question

prove that b2+c2/b+c + c2+a2/c+a + a2+b2/a+b > a+b+c
​

Answer 1

Answer:

Answer

∣

∣

∣

∣

∣

∣

∣

∣

(b+c)

2

b

2

c

2

a

2

(c+a)

2

c

2

a

2

b

2

(a+b)

2

∣

∣

∣

∣

∣

∣

∣

∣

=(b+c)

2

[(c+a)

2

×(a+b)

2

−b

2

×c

2

]−

a

2

[b

2

×(a+b)

2

−b

2

×c

2

]+a

2

[b

2

c

2

−c

2

(c+a)

2

]

=b

2

+c

2

+2ab[(c

2

+a

2

+2ac)×(a

2

+b

2

+2ab)−b

2

c

2

]−a

2

[b

2

×(a

2

+b

2

+2ab)−b

2

c

2

]+a

2

[b

2

c

2

−c

2

(c

2

+a

2

+2ac)]

onsolvingandtakingcomonweget

=2abc(a

2

+b

2

+c

2

+2ab+2bc+2ca)

2

=2abc(a+b+c)

2

proved