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Question Number 20631 by Tinkutara last updated on 30/Aug/17
Solve the inequality (x + 3)^5 − (x − 1)^5 ≥ 244.
$${Solve}\:{the}\:{inequality} \\ $$$$\left({x}\:+\:\mathrm{3}\right)^{\mathrm{5}} \:−\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{5}} \:\geqslant\:\mathrm{244}. \\ $$
Answered by mrW1 last updated on 30/Aug/17
let u=x+1 (u+2)^5 −(u−2)^5 ≥244 u^5 +5×2u^4 +10×4u^3 +10×8u^2 +5×16u+32 −u^5 +5×2u^4 −10×4u^3 +10×8u^2 −5×16u+32 =2×(5×2u^4 +10×8u^2 +32)≥244 u^4 +8u^2 −9≥0 (u^2 +9)(u^2 −1)≥0 u^2 −1≥0 u^2 ≥1 u≥1 or u≤−1 x+1≥1 or x+1≤−1 ⇒x≥0 or x≤−2
$$\mathrm{let}\:\mathrm{u}=\mathrm{x}+\mathrm{1} \\ $$$$\left(\mathrm{u}+\mathrm{2}\right)^{\mathrm{5}} −\left(\mathrm{u}−\mathrm{2}\right)^{\mathrm{5}} \geqslant\mathrm{244} \\ $$$$\mathrm{u}^{\mathrm{5}} +\mathrm{5}×\mathrm{2u}^{\mathrm{4}} +\mathrm{10}×\mathrm{4u}^{\mathrm{3}} +\mathrm{10}×\mathrm{8u}^{\mathrm{2}} +\mathrm{5}×\mathrm{16u}+\mathrm{32} \\ $$$$−\mathrm{u}^{\mathrm{5}} +\mathrm{5}×\mathrm{2u}^{\mathrm{4}} −\mathrm{10}×\mathrm{4u}^{\mathrm{3}} +\mathrm{10}×\mathrm{8u}^{\mathrm{2}} −\mathrm{5}×\mathrm{16u}+\mathrm{32} \\ $$$$=\mathrm{2}×\left(\mathrm{5}×\mathrm{2u}^{\mathrm{4}} +\mathrm{10}×\mathrm{8u}^{\mathrm{2}} +\mathrm{32}\right)\geqslant\mathrm{244} \\ $$$$\mathrm{u}^{\mathrm{4}} +\mathrm{8u}^{\mathrm{2}} −\mathrm{9}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{u}^{\mathrm{2}} +\mathrm{9}\right)\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{1}\geqslant\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{2}} \geqslant\mathrm{1} \\ $$$$\mathrm{u}\geqslant\mathrm{1}\:\mathrm{or}\:\mathrm{u}\leqslant−\mathrm{1} \\ $$$$\mathrm{x}+\mathrm{1}\geqslant\mathrm{1}\:\mathrm{or}\:\mathrm{x}+\mathrm{1}\leqslant−\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\geqslant\mathrm{0}\:\mathrm{or}\:\mathrm{x}\leqslant−\mathrm{2} \\ $$
Commented by Tinkutara last updated on 30/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 30/Aug/17
let x+1=y ; then (y+2)^5 −(y−2)^5 ≥ 244 ⇒ 10y^4 +80y^2 +32 ≥ 122 y^4 +8y^2 ≥ 9 (y^2 +4)^2 ≥ 25 ⇒ y^2 ≥ 1 or (x+1)^2 ≥ 1 (x+2)x ≥ 0 x∈ (−∞, −2] ∪ [0, ∞) .
$${let}\:{x}+\mathrm{1}={y}\:;\:{then} \\ $$$$\left({y}+\mathrm{2}\right)^{\mathrm{5}} −\left({y}−\mathrm{2}\right)^{\mathrm{5}} \:\geqslant\:\mathrm{244} \\ $$$$\Rightarrow\:\mathrm{10}{y}^{\mathrm{4}} +\mathrm{80}{y}^{\mathrm{2}} +\mathrm{32}\:\geqslant\:\mathrm{122} \\ $$$$\:\:\:\:\:\:{y}^{\mathrm{4}} +\mathrm{8}{y}^{\mathrm{2}} \geqslant\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{25} \\ $$$$\Rightarrow\:\:\:\:{y}^{\mathrm{2}} \:\:\geqslant\:\mathrm{1}\:\:{or}\:\:\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\left({x}+\mathrm{2}\right){x}\:\geqslant\:\mathrm{0} \\ $$$$\:\:{x}\in\:\left(−\infty,\:−\mathrm{2}\right]\:\cup\:\left[\mathrm{0},\:\infty\right)\:. \\ $$
Commented by Tinkutara last updated on 30/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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