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

$$\mathrm{Can}\mathrm{we}\mathrm{show}\mathrm{a}+\mathrm{b}<{\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2$$ $$\mathrm{\forall }\mathrm{a},\mathrm{b}\in \mathbb{N}$$

Commented byFrix last updated on 08/Dec/22

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

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

$$fora,b>2wecanprove.$$ $$ifa=b:$$ $$a^2-ab+b^2=a^2>2a=a+b✓$$ $$ifa\ne b,saya>b:$$ $$a^2-ab+b^2=a(a-b)+b^2⩾a+b^2>a+b✓$$

Commented byRasheed.Sindhi last updated on 09/Dec/22

$$\mathcal{T}\mathit{h}\mathit{a}\mathit{n}\mathcal{X}\mathcal{S}\mathit{i}\mathit{r}!$$

Answered by manxsol last updated on 09/Dec/22

$$analisis$$ $$a=b\Rightarrow a^2-a.a+a^2⟩a+a$$ $$a^2⟩2a\Rightarrow a(a-2)⟩0$$ $$aϵ⟨-\propto ,0⟩\cup ⟨2,0⟩$$ $$a⟩2$$ $$ifa\ne banda⟩b$$ $$propertie:a+b⟩2\sqrt{ab}$$ $$a^3+b^3⟩2\sqrt{a^3{b}^3}$$ $$a^3+b^3⟩2ab\sqrt{ab}I$$ $${(a+b)}^2⟩2ab⟩2\sqrt{ab}II$$ $$I/II$$ $$\frac{a^3+b^3}{{(a+b)}^2}⟩\frac{2ab\sqrt{ab}}{2\sqrt{ab}}$$ $$a^3+b^3⟩{(a+b)}^{2}ab$$ $$a^2-ab+b^2⟩(a+b)ab⟩a+b$$ $$a^2-ab+b^2⟩a+b$$

Commented bymanxsol last updated on 09/Dec/22

$$Error,Iammissingthe$$ $$analysisthatyoudid.$$ $$ThankSirW.$$

Commented byRasheed.Sindhi last updated on 09/Dec/22

$$\mathcal{T}\mathit{h}\mathit{a}\mathit{n}\mathcal{X}\mathcal{S}\mathit{i}\mathit{r}!$$

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