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Question Number 144537 by loveineq last updated on 26/Jun/21

Let a,b>0 and a+b = 2. Prove that (1) ((a^3 +b^3 )/2)−2(1−ab) ≥ 1 (2) ((a^2 +b^2 )/2)−2(1−ab) ≤ 1

$$\mathrm{Let}\:{a},{b}>\mathrm{0}\:\mathrm{and}\:{a}+{b}\:=\:\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\:\:\:\:\: \\ $$ $$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right)\:\geqslant\:\mathrm{1} \\ $$ $$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right)\:\leqslant\:\mathrm{1} \\ $$

Answered by mnjuly1970 last updated on 26/Jun/21

(1) : (((a+b)(a^( 2) +b^( 2) −ab))/2)−2+2ab =a^( 2) +b^2 −2+ab=(a+b)^2 −ab−2 =2−ab ≥_(inequality) ^(am −gm) 2−((a+b)/2) =1 ...

$$\left(\mathrm{1}\right)\::\:\:\frac{\left({a}+{b}\right)\left({a}^{\:\mathrm{2}} +{b}^{\:\mathrm{2}} −{ab}\right)}{\mathrm{2}}−\mathrm{2}+\mathrm{2}{ab} \\ $$ $$\:\:\:\:\:\:\:\:={a}^{\:\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}+{ab}=\left({a}+{b}\right)^{\mathrm{2}} −{ab}−\mathrm{2} \\ $$ $$\:\:=\mathrm{2}−{ab}\:\underset{{inequality}} {\overset{{am}\:−{gm}} {\geqslant}}\mathrm{2}−\frac{{a}+{b}}{\mathrm{2}}\:=\mathrm{1}\:... \\ $$

Commented byloveineq last updated on 26/Jun/21

thanks

$$ \\ $$ $${thanks} \\ $$

Answered by Olaf_Thorendsen last updated on 26/Jun/21

(1) : x = ((a^3 +b^3 )/2)−2(1−ab) x = (((a+b)^3 −3ab(a+b))/2)−2(1−ab) x = ((8−6ab)/2)−2(1−ab) x = 2−ab = 2−a(2−a) x = a^2 −2a+2 = (a−1)^2 +1 ≥ 1 (2) y = ((a^2 +b^2 )/2)−2(1−ab) y = (((a+b)−2ab)/2)−2(1−ab) y = ((2−2ab)/2)−2(1−ab) y = ab−1 = a(2−a)−1 y = −a^2 +2a−1 = −(a−1)^2 ≤ 0 ≤ 1

$$\left(\mathrm{1}\right)\::\:{x}\:=\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right) \\ $$ $${x}\:=\:\frac{\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right)}{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right) \\ $$ $${x}\:=\:\frac{\mathrm{8}−\mathrm{6}{ab}}{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right) \\ $$ $${x}\:=\:\mathrm{2}−{ab}\:=\:\mathrm{2}−{a}\left(\mathrm{2}−{a}\right) \\ $$ $${x}\:=\:{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{2}\:=\:\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\:\geqslant\:\mathrm{1} \\ $$ $$\left(\mathrm{2}\right)\:{y}\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right) \\ $$ $${y}\:=\:\frac{\left({a}+{b}\right)−\mathrm{2}{ab}}{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right) \\ $$ $${y}\:=\:\frac{\mathrm{2}−\mathrm{2}{ab}}{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−{ab}\right) \\ $$ $${y}\:=\:{ab}−\mathrm{1}\:=\:{a}\left(\mathrm{2}−{a}\right)−\mathrm{1} \\ $$ $${y}\:=\:−{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{1}\:=\:−\left({a}−\mathrm{1}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{0}\:\leqslant\:\mathrm{1} \\ $$

Commented byloveineq last updated on 26/Jun/21

thanks

$$ \\ $$ $${thanks} \\ $$

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