parmenides51 wrote: Let $-1 \le x, y \le 1$. Prove the inequality $$2\sqrt{(1- x^2)(1 - y^2) } \le 2(1 - x)(1 - y) + 1 $$

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$\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ It screams for trigonometry. There exist $\alpha,\beta\in(0^\circ;90^\circ)$ such that $$x=\cos2\alpha,\ y=\cos2\beta.$$Inequality becomes $$2\sin2\alpha\sin2\beta\le8\sin^2\alpha\sin^2\beta+1,$$$$8\sin\alpha\sin\beta\cos(\alpha+\beta)\le 1.$$Since $2\sin\alpha\cos(\alpha+\beta)=\sin(2\alpha+\beta)-\sin\beta$ we have by AM-GM $$8\sin\alpha\sin\beta\cos(\alpha+\beta)= 4\sin\beta\left(\sin(2\alpha+\beta)-\sin\beta\right)\le \left(\sin\beta+\left(\sin(2\alpha+\beta)-\sin\beta\right)\right)^2=\sin(2\alpha+\beta)^2\le 1.$$Equality iff $$x=\frac12=y.\blacksquare$$#1769

HEY! on substituting x=sin@ y=cos@ then on substituting and applying sum to product transformations 2cos(@/2)<=3 on computing there actually exists no inequality

parmenides51 wrote: Let $-1 \le x, y \le 1$. Prove the inequality $$2\sqrt{(1- x^2)(1 - y^2) } \le 2(1 - x)(1 - y) + 1 $$

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Let $-1 \le x, y \le 1$. Prove that $$-4\leq \sqrt{(1- x^2)(1 - y^2) }-(1 - x)(1 - y) \le \frac{1}{2} $$$$-6\leq \sqrt{(1- x^2)(1 - y^2) }-(1 - x)(1 -2 y) \le 2$$$$-9\leq \sqrt{(1- x^2)(1 - y^2) }-(1 - 2x)(1 - 2y) \le 3$$Let $-1 \le x, y,z \le 1$. Prove that $$-27\leq \sqrt{(1- x^2)(1- y^2)(1-z^2)}- (1 -2 x)(1 - 2y)(1 -2 z) \le 9$$$$-2\leq \sqrt{(1- x^2)(1- y^2)(1-z^2)}- (1 - x)(1 - 2y)(1 -2 z) \le 6$$$$-12\leq \sqrt{(1- x^2)(1- y^2)(1-z^2)}- (1 - x)(1 - 2y)(1 - z) \le 4$$

solved also here and here (as p1)

Let $-1 \le x, y \le 1$. Prove that $$ \sqrt{(1- x)(1 - y) }-(1 - x^2)(1 - y^2) \le 2 $$$$\sqrt{(1- x)(1 - y) }-(1 - 2x^2)(1 - 2y^2) \geq -1$$$$\sqrt{(1- x)(1 - y) }-(1 - x^2)(1 - 2y^2) \leq\frac{1}{8}(9+5\sqrt 5)$$$$-\frac{3}{4}\leq\sqrt{(1- 2x)(1 - 2y) }-(1 -x^2)(1 - y^2) \leq 3$$$$-\frac{3}{4}\leq\sqrt{(1- x)(1 - 2y) }-(1 -x^2)(1 - y^2) \leq \sqrt 6$$$$-\frac{3}{4}\leq\sqrt{(1- x)(1 - 2y) }-(1 -2x^2)(1 - y^2) \leq\frac{11}{4}$$

sqing wrote: Let $-1 \le x, y \le 1$. Prove that $$ \sqrt{(1- x)(1 - y) }-(1 - x^2)(1 - y^2) \le 2 $$$$\sqrt{(1- x)(1 - y) }-(1 - 2x^2)(1 - 2y^2) \geq -1$$

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