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Question Number 142986 by mnjuly1970 last updated on 08/Jun/21

Commented by MJS_new last updated on 08/Jun/21

$if y = x the expression is constantly \sqrt[3]{4}$
Commented by mnjuly1970 last updated on 08/Jun/21

$p e r f e c . . .$
	Answered by TheHoneyCat last updated on 17/Jun/21
	$ler c: { (R,→,R),(x,→,x^3 ) :} ∈C^∞ (R) which, by checking the derrivative, is obviously a bijection So: Min_(x,y>1) {((^3 (√(x−1))+^3 (√(y−1)))/(^3 (√(x+y−2))))} =Min_(x,y>0) {((^3 (√x)+^3 (√y))/(^3 (√(x+y))))} _((x⇋x−1, y⇋y−1)) =c^(−1) (Min_(x,y>0) {(((^3 (√x)+^3 (√y))^3 )/(x+y))}) =c^(−1) (Min_(x,y>0) {(((x+y)^3 )/(x^3 +y^3 ))}) _((x⇋^3 (√x), y⇋^3 (√y))) =c^(−1) (Min_(x,y>0) {((x^3 +y^3 +6xy^2 +6yx^2 )/(x^3 +y^3 ))}) =c^(−1) (Min_(x,y>0) {1+6((xy^2 +yx^2 )/(x^3 +y^3 ))}) =c^(−1) (1+6×Min_(x,y>0) {((xy^2 +yx^2 )/(x^3 +y^3 ))}) let (Un)=(1,(1/n))_(n∈N) (Un)∈{x,y ∣ x>0 and y>0} = ′′x,y>0′′ and (((1/n^2 )+(1/n))/(1+(1/n^3 )))→_(n→+∞) 0 So, Min_(x,y>0) {((xy^2 +yx^2 )/(x^3 +y^3 ))}≤0 But ∀(x,y)∈(R_+ ^∗ )^2 ((xy^2 +yx^2 )/(x^3 +y^3 ))}≥0 (∗) So, Min_(x,y>0) {((xy^2 +yx^2 )/(x^3 +y^3 ))}=0 And thus : Min_(x,y>1) {((^3 (√(x−1))+^3 (√(y−1)))/(^3 (√(x+y−2))))}=c^(−1) (1)=1 Your minimum is worth 1 By the way, it is possible to proove that this value cannot be obtained with (x,y)∈(R_+ ^∗ )^2 (∗) is in fact a strict innequality so technically it is not called a Max but a Sup and (∗) would not contradict anything.$
	$ler c : {\begin{cases} R & \to & R \\ x & \to & x^{3} \end{cases} \in C^{\infty} (R) which, by checking the derrivative, is obviously a bijection$ $So :$ ${Min}_{x, y > 1} {\frac{^{3} \sqrt{x - 1} +^{3} \sqrt{y - 1}}{^{3} \sqrt{x + y - 2}}}$ $= {Min}_{x, y > 0} {\frac{^{3} \sqrt{x} +^{3} \sqrt{y}}{^{3} \sqrt{x + y}}}_{(x ⇋ x - 1, y ⇋ y - 1)}$ $= c^{- 1} ({Min}_{x, y > 0} {\frac{{(^{3} \sqrt{x} +^{3} \sqrt{y})}^{3}}{x + y}})$ $= c^{- 1} ({Min}_{x, y > 0} {\frac{{(x + y)}^{3}}{x^{3} + y^{3}}})_{(x ⇋^{3} \sqrt{x}, y ⇋^{3} \sqrt{y})}$ $= c^{- 1} ({Min}_{x, y > 0} {\frac{x^{3} + y^{3} + 6 {x y}^{2} + 6 {y x}^{2}}{x^{3} + y^{3}}})$ $= c^{- 1} ({Min}_{x, y > 0} {1 + 6 \frac{{x y}^{2} + {y x}^{2}}{x^{3} + y^{3}}})$ $= c^{- 1} (1 + 6 \times {Min}_{x, y > 0} {\frac{{x y}^{2} + {y x}^{2}}{x^{3} + y^{3}}})$ $let (U n) = {(1, \frac{1}{n})}_{n \in N}$ $(U n) \in {x, y ∣ x > 0 and y > 0} =^{″} x, y > 0^{″}$ $and \frac{\frac{1}{n^{2}} + \frac{1}{n}}{1 + \frac{1}{n^{3}}} \underset{n \to + \infty}{\to} 0$ $So, {Min}_{x, y > 0} {\frac{{x y}^{2} + {y x}^{2}}{x^{3} + y^{3}}} ⩽ 0$ $But \forall (x, y) \in {(R_{+}^{})}^{2} \frac{{x y}^{2} + {y x}^{2}}{x^{3} + y^{3}}} ⩾ 0 ()$ $So, {Min}_{x, y > 0} {\frac{{x y}^{2} + {y x}^{2}}{x^{3} + y^{3}}} = 0$ $And thus :$ ${Min}_{x, y > 1} {\frac{^{3} \sqrt{x - 1} +^{3} \sqrt{y - 1}}{^{3} \sqrt{x + y - 2}}} = c^{- 1} (1) = 1$ $Your minimum is worth 1$ $B y t h e w a y, i t i s p o s s i b l e t o p r o o v e t h a t t h i s v a l u e c a n n o t b e o b t a i n e d w i t h (x, y) \in {(R_{+}^{})}^{2}$ $() i s i n f a c t a s t r i c t i n n e q u a l i t y$ $s o t e c h n i c a l l y i t i s n o t c a l l e d a M a x b u t a S u p$ $a n d (*) w o u l d n o t c o n t r a d i c t a n y t h i n g .$
	Answered by ajfour last updated on 17/Jun/21

	$x - 1 = p^{3}, y - 1 = q^{3}$ $z = \frac{p + q}{{(p^{3} + q^{3})}^{1 / 3}} = \frac{m + 1}{{(m^{3} + 1)}^{1 / 3}}$ $z = \frac{{(m + 1)}^{2 / 3}}{{(m^{2} - m + 1)}^{1 / 3}}$ $z^{3} = \frac{{(m + 1)}^{2}}{m^{2} - m + 1} = \frac{{(m + 1)}^{2}}{{(m + 1)}^{2} - 3 (m + 1) + 3}$ $l e t \frac{m + 1}{\sqrt{3}} = t$ $\Rightarrow z^{3} = \frac{3 t^{2}}{3 t^{2} - 3 \sqrt{3} t + 3}$ $z^{3} = \frac{t}{t + \frac{1}{t} - \sqrt{3}}$ $s a y t + \frac{1}{t} = s \Rightarrow t = \frac{s}{2} \pm \sqrt{\frac{s^{2}}{4} - 1}$ $z^{3} = \frac{s - \sqrt{s^{2} - 4}}{2 (s - \sqrt{3})}$ $\frac{d (2 z^{3})}{d s} = \frac{(s - \sqrt{3}) (1 - \frac{s}{\sqrt{s^{2} - 4}}) - (s - \sqrt{s^{2} - 4})}{{(s - \sqrt{3})}^{2}}$ $\frac{d (2 z^{3})}{d s} = \frac{\frac{\sqrt{3} s - s^{2}}{\sqrt{s^{2} - 4}} - \sqrt{3} + \sqrt{s^{2} - 4}}{{(s - 3 \sqrt{3})}^{2}}$ $= \frac{\sqrt{3} s - 4 - \sqrt{3} \sqrt{s^{2} - 4}}{\sqrt{s^{2} - 4} {(s - 3 \sqrt{3})}^{2}}$ $l e t {(\sqrt{3} s - 4)}^{2} = 3 (s^{2} - 4)$ $\Rightarrow s = \frac{7}{2 \sqrt{3}}$ $z^{3} = \frac{(\frac{7 - 1}{2 \sqrt{3}})}{(\frac{2}{2 \sqrt{3}})} = 3$ $z = \sqrt[3]{3}$
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