Can-we-show-a-b-lt-a-2-ab-b-2-a-b-N-

Question Number 182379 by Rasheed.Sindhi last updated on 08/Dec/22

Can we show a + b < a^{2} - ab + b^{2}

\forall a, b \in N

Commented by Frix last updated on 08/Dec/22

Check it for a \in {0, 1, 2}

Answered by mr W last updated on 08/Dec/22

f o r a, b > 2 w e c a n p r o v e .

i f a = b :

a^{2} - a b + b^{2} = a^{2} > 2 a = a + b ✓

i f a \neq b, s a y a > b :

a^{2} - a b + b^{2} = a (a - b) + b^{2} ⩾ a + b^{2} > a + b ✓

Commented by Rasheed.Sindhi last updated on 09/Dec/22

T h a n X S i r!

Answered by manxsol last updated on 09/Dec/22

a n a l i s i s

a = b \Rightarrow a^{2} - a . a + a^{2} ⟩ a + a

a^{2} ⟩ 2 a \Rightarrow a (a - 2) ⟩ 0

a ϵ ⟨ - \propto, 0 ⟩ \cup ⟨ 2, 0 ⟩

a ⟩ 2

i f a \neq b a n d a ⟩ b

p r o p e r t i e : a + b ⟩ 2 \sqrt{a b}

a^{3} + b^{3} ⟩ 2 \sqrt{a^{3} b^{3}}

a^{3} + b^{3} ⟩ 2 a b \sqrt{a b} I

{(a + b)}^{2} ⟩ 2 a b ⟩ 2 \sqrt{a b} I I

I / I I

\frac{a^{3} + b^{3}}{{(a + b)}^{2}} ⟩ \frac{2 a b \sqrt{a b}}{2 \sqrt{a b}}

a^{3} + b^{3} ⟩ {(a + b)}^{2} a b

a^{2} - a b + b^{2} ⟩ (a + b) a b ⟩ a + b

a^{2} - a b + b^{2} ⟩ a + b

Commented by manxsol last updated on 09/Dec/22

E r r o r, I a m m i s s i n g t h e

a n a l y s i s t h a t y o u d i d .

T h a n k S i r W .

Commented by Rasheed.Sindhi last updated on 09/Dec/22

T h a n X S i r!