Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Can we show a+b<a^2 −ab+b^2 ∀ a,b∈N

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

Commented byFrix last updated on 08/Dec/22

Check it for a∈{0, 1, 2}

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

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

for a,b>2 we can prove. if a=b: a^2 −ab+b^2 =a^2 >2a=a+b ✓ if a≠b, say a>b: a^2 −ab+b^2 =a(a−b)+b^2 ≥a+b^2 >a+b ✓

$${for}\:{a},{b}>\mathrm{2}\:{we}\:{can}\:{prove}. \\ $$ $${if}\:{a}={b}: \\ $$ $${a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} ={a}^{\mathrm{2}} >\mathrm{2}{a}={a}+{b}\:\checkmark \\ $$ $${if}\:{a}\neq{b},\:{say}\:{a}>{b}: \\ $$ $${a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} ={a}\left({a}−{b}\right)+{b}^{\mathrm{2}} \geqslant{a}+{b}^{\mathrm{2}} >{a}+{b}\:\checkmark \\ $$

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

ThanX Sir!

$$\boldsymbol{\mathcal{T}{han}\mathcal{X}}\:\boldsymbol{\mathcal{S}{ir}}! \\ $$

Answered by manxsol last updated on 09/Dec/22

analisis a=b⇒a^2 −a.a+a^2 ⟩a+a a^2 ⟩2a⇒a(a−2)⟩0 aε⟨−∝,0⟩∪⟨2,0⟩ a⟩2 if a≠b and a⟩b propertie : a+b⟩2(√(ab)) a^3 +b^3 ⟩2(√(a^3 b^3 )) a^3 +b^3 ⟩2ab(√(ab)) I (a+b)^2 ⟩2ab⟩2(√(ab)) II I/II ((a^3 +b^3 )/((a+b)^2 )) ⟩ ((2ab(√(ab)))/( 2(√(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

$${analisis} \\ $$ $${a}={b}\Rightarrow{a}^{\mathrm{2}} −{a}.{a}+{a}^{\mathrm{2}} \rangle{a}+{a} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} \rangle\mathrm{2}{a}\Rightarrow{a}\left({a}−\mathrm{2}\right)\rangle\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:{a}\epsilon\langle−\propto,\mathrm{0}\rangle\cup\langle\mathrm{2},\mathrm{0}\rangle \\ $$ $${a}\rangle\mathrm{2} \\ $$ $${if}\:{a}\neq{b}\:\:{and}\:{a}\rangle{b} \\ $$ $${propertie}\:\:\:\::\:\:{a}+{b}\rangle\mathrm{2}\sqrt{{ab}}\:\:\:\:\:\: \\ $$ $$\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \:\rangle\mathrm{2}\sqrt{{a}^{\mathrm{3}} {b}^{\mathrm{3}} } \\ $$ $$\:\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \rangle\mathrm{2}{ab}\sqrt{{ab}}\:\:\:\:\:{I} \\ $$ $$\left({a}+{b}\right)^{\mathrm{2}} \rangle\mathrm{2}{ab}\rangle\mathrm{2}\sqrt{{ab}}\:\:\:\:\:{II} \\ $$ $${I}/{II} \\ $$ $$\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{\left({a}+{b}\right)^{\mathrm{2}} }\:\:\rangle\:\frac{\mathrm{2}{ab}\sqrt{{ab}}}{\:\mathrm{2}\sqrt{{ab}}} \\ $$ $$\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \rangle\left({a}+{b}\right)^{\mathrm{2}} {ab} \\ $$ $${a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \rangle\left({a}+{b}\right){ab}\rangle\:\:{a}+{b} \\ $$ $${a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \rangle\:{a}+{b} \\ $$

Commented bymanxsol last updated on 09/Dec/22

Error,I am missing the analysis that you did. Thank Sir W.

$${Error},{I}\:{am}\:{missing}\:{the} \\ $$ $$\:{analysis}\:{that}\:{you}\:{did}. \\ $$ $${Thank}\:{Sir}\:{W}. \\ $$

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

ThanX Sir!

$$\boldsymbol{\mathcal{T}{han}\mathcal{X}}\:\boldsymbol{\mathcal{S}{ir}}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com