Menu Close

Show-that-a-3-b-3-a-2-b-2-gt-a-2-b-2-a-b-

Tinku Tara 0 Comments
Pin




Question Number 202307 by MATHEMATICSAM last updated on 24/Dec/23
Show that ((a^3 + b^3 )/(a^2 + b^2 )) > ((a^2 + b^2 )/(a + b))
$$\mathrm{Show}\:\mathrm{that}\:\frac{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:>\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }{{a}\:+\:{b}} \\ $$
Answered by aleks041103 last updated on 24/Dec/23
(a^3 +b^3 )(a+b)−(a^2 +b^2 )^2 = =a^4 +a^3 b+ab^3 +b^4 −a^4 −b^4 −2a^2 b^2 = =ab(a^2 +b^2 −2ab)=ab(a−b)^2 ⇒ a,b>0 ∧ a≠b ⇒ ((a^3 +b^3 )/(a^2 +b^2 ))>((a^2 +b^2 )/(a+b))
$$\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)\left({a}+{b}\right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} = \\ $$$$={a}^{\mathrm{4}} +{a}^{\mathrm{3}} {b}+{ab}^{\mathrm{3}} +{b}^{\mathrm{4}} −{a}^{\mathrm{4}} −{b}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} = \\ $$$$={ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right)={ab}\left({a}−{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${a},{b}>\mathrm{0}\:\wedge\:{a}\neq{b}\:\Rightarrow\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }>\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}} \\ $$
Answered by MM42 last updated on 24/Dec/23
if a=−1 & b=2 ⇒(7/5)>5 its wrong.
$${if}\:\:\:{a}=−\mathrm{1}\:\&\:\:{b}=\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{7}}{\mathrm{5}}>\mathrm{5}\:\:\:{its}\:\:{wrong}. \\ $$$$ \\ $$
Commented by aleks041103 last updated on 24/Dec/23
yes. for a and b different positive numbers the inequality is true
$${yes}.\: \\ $$$${for}\:{a}\:{and}\:{b}\:{different}\:{positive}\:{numbers} \\ $$$${the}\:{inequality}\:{is}\:{true} \\ $$
Answered by Frix last updated on 24/Dec/23
((a^3 +b^3 )/(a^2 +b^2 ))−((a^2 +b^2 )/(a+b))>0 ((a(a−b)^2 b)/((a+b)(a^2 +b^2 )))>0 ((ab)/(a+b))>0 True if ab>0∧a+b>0 ∨ ab<0∧a+b<0 ⇔ a>0∧b>0 ∨ a<0∧b<0
$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{a}+{b}}>\mathrm{0} \\ $$$$\frac{{a}\left({a}−{b}\right)^{\mathrm{2}} {b}}{\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}>\mathrm{0} \\ $$$$\frac{{ab}}{{a}+{b}}>\mathrm{0} \\ $$$$\mathrm{True}\:\mathrm{if} \\ $$$${ab}>\mathrm{0}\wedge{a}+{b}>\mathrm{0}\:\vee\:{ab}<\mathrm{0}\wedge{a}+{b}<\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${a}>\mathrm{0}\wedge{b}>\mathrm{0}\:\vee\:{a}<\mathrm{0}\wedge{b}<\mathrm{0} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *