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Let $a,b,c,d$ be real numbers such that $a^2+b^2=c^2+d^2=1$. Find the maximum value of $(1-a)(1-c)+(1-b)(1-d)$.


I tried substituting $a=\sin x, b =\cos x, c = \sin y, d=\cos y$, then expanded $(1-a)(1-c)+(1-b)(1-d)$. However this just leads to an ugly expression, and I'm not sure how to proceed

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    $\begingroup$ What about lagrange multipliers$?$ Have you tried that method$?$ $\endgroup$ Feb 18, 2023 at 17:16
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    $\begingroup$ I haven't learned Calculus yet :( $\endgroup$ Feb 18, 2023 at 17:17
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    $\begingroup$ Notice that : $$(1 - a) (1 - c) + (1 - b) (1 - d) = (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{w})$$ with $\vec{u}(1, 1)$, $\vec{v}(a, b)$ and $\vec{w}(c, d)$. $\endgroup$
    – Essaidi
    Feb 18, 2023 at 17:52
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    $\begingroup$ By the symmetry of the expression, we must have $a = b = c = d$. By the minus sign, $a, b, c$ and $d$ must be $\leq 0$. So $a = b = c = d = -\dfrac{\sqrt{2}}{2}$. $\endgroup$
    – Essaidi
    Feb 18, 2023 at 17:57
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    $\begingroup$ @Essaidi Please do not assume that due to symmetry, extrema must occur when all terms are equal. $\endgroup$
    – Calvin Lin
    Feb 19, 2023 at 0:19

2 Answers 2

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By the Cauchy-Schwarz inequality, we have: $$1 \times 1=(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 \implies 1 \geq ac+bd;$$

and, $$(a^2+b^2)(1^2+1^2)\geq (a+b)^2 \implies \sqrt 2 \geq -a-b, $$ similarly, $\sqrt 2 \geq -c-d$.

Therefore,

$$(1-a)(1-c)+(1-b)(1-d) \\ =2+(-a-b)+(-c-d)+(ac+bd) \\ \leq 2+2\sqrt 2+1=3+2\sqrt 2.$$

The equality case happens at $a=b=c=d=-\frac{\sqrt 2}{2}.$

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    $\begingroup$ Line $\ \ \ldots \Rightarrow \sqrt 2 \ldots\ \ $ is not clear. $\endgroup$
    – Wlod AA
    Feb 18, 2023 at 18:07
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    $\begingroup$ @WlodAA $A^2+B^2 = 1$, so take the square root. and chhose $-a-b$ because in the last simplifications, that yields the maximal value $\endgroup$
    – D S
    Feb 18, 2023 at 18:13
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WLOG, Let $a=-\sin2x,c=-\sin2y, b=-\cos2x,d=-\cos2y$

$$S=(1-a)(1-c)+(1-b)(1-d)$$

$$=2+(\sin2x+\sin2y+\cos2x+\cos2y)+\cos2(x-y)$$

$$=1+2\cos(x-y)(\sin(x+y)+\cos(x+y))+2\cos^2(x-y)$$ will be maximum

$(1)$ if $\cos(x-y)$ is maximum and so is $\sin(x+y)+\cos(x+y)$

$\cos(x-y)$ will be maximum $(=1)$ which needs $x=2n\pi+y$

$\sin(x+y)+\cos(x+y)$ reduces to $\sin2y+\cos2y=\sqrt2\cos\left(2y-\dfrac\pi4\right)$

which is $\le\sqrt2$ the equality occurs if $2y=2m\pi+\dfrac\pi4$

So, $$S\le1+2\cdot1\cdot\sqrt2+2\cdot1$$

$(2)$ or if $\cos(x-y)$ is minimum and $\cos(x+y)+\sin(x+y)$ is minimum i.e., $-\sqrt2$

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