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Question Number 199468 by Calculusboy last updated on 04/Nov/23

	Answered by witcher3 last updated on 04/Nov/23
	$(x/(2021π))⇔ ∫_0 ^1 (1/e^t^(2021) )(1+t^(2022) )^(1/(2022)) dt=I ∫_0 ^1 e^(−t^(2021) ) (1+t^(2022) )^(1/(2022)) dt e^x ≥1+x⇒e^(−x) ≤(1/(1+x)),∀x≥0 prof x→e^x is convexe graphe of x→e^x is up to tangent in/zero e^x ≥e^0 (x−0)+e^0 =x+1 (1+t)^a ≤1+at,∀a∈[0,1[ t>0 t→(1+t)^a concave ⇒(1+t)^a ≤at+1 {tangent in zero} e^(−t^(2021) ) ≤(1/(1+t^(2021) )) (1+t^(2022) )^(1/(2022)) ≤1+(t^(2022) /(2022))<1+(t^(2021) /(2022))<1+t^(2021) ,∀t∈[0,1] I≤∫_0 ^1 ((1+(t^(2022) /(2022)))/(1+t^(2021) ))dt<∫_0 ^1 ((1+t^(2021) )/(1+t^(2021) ))dt=1 ∫_0 ^(2021π) (1/e^(((x/(2021π)))^(2021) ) )(((1+((x/(2021π)))^(2022) ))^(1/(2022)) dx<1$
	$\frac{x}{2021 π} \Leftrightarrow$ $\int_{0}^{1} \frac{1}{e^{t^{2021}}} {(1 + t^{2022})}^{\frac{1}{2022}} dt = I$ $\int_{0}^{1} e^{- t^{2021}} {(1 + t^{2022})}^{\frac{1}{2022}} dt$ $e^{x} ⩾ 1 + x \Rightarrow e^{- x} ⩽ \frac{1}{1 + x}, \forall x ⩾ 0$ $prof x \to e^{x} is convexe graphe of x \to e^{x} is up to tangent in / zero$ $e^{x} ⩾ e^{0} (x - 0) + e^{0} = x + 1$ ${(1 + t)}^{a} ⩽ 1 + at, \forall a \in [0, 1 [t > 0$ $t \to {(1 + t)}^{a} concave \Rightarrow {(1 + t)}^{a} ⩽ at + 1 {tangent in zero}$ $e^{- t^{2021}} ⩽ \frac{1}{1 + t^{2021}}$ ${(1 + t^{2022})}^{\frac{1}{2022}} ⩽ 1 + \frac{t^{2022}}{2022} < 1 + \frac{t^{2021}}{2022} < 1 + t^{2021}, \forall t \in [0, 1]$ $I ⩽ \int_{0}^{1} \frac{1 + \frac{t^{2022}}{2022}}{1 + t^{2021}} dt < \int_{0}^{1} \frac{1 + t^{2021}}{1 + t^{2021}} dt = 1$ $\int_{0}^{2021 π} \frac{1}{e^{{(\frac{x}{2021 π})}^{2021}}} \sqrt[2022]{(1 + {(\frac{x}{2021 π})}^{2022}} dx < 1$
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