# Maximize $(1-a)(1-c)+(1-b)(1-d)$ over $a^2+b^2=c^2+d^2=1$. [closed]

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

• What about lagrange multipliers$?$ Have you tried that method$?$ Feb 18, 2023 at 17:16
• I haven't learned Calculus yet :( Feb 18, 2023 at 17:17
• 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 $\vec{w}(c, d)$. Feb 18, 2023 at 17:52
• 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 $a = b = c = d = -\dfrac{\sqrt{2}}{2}$. Feb 18, 2023 at 17:57
• @Essaidi Please do not assume that due to symmetry, extrema must occur when all terms are equal. Feb 19, 2023 at 0:19

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 $$a=b=c=d=-\frac{\sqrt 2}{2}.$$

• Line $\ \ \ldots \Rightarrow \sqrt 2 \ldots\ \$ is not clear. Feb 18, 2023 at 18:07
• @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
– D S
Feb 18, 2023 at 18:13

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., $$-\sqrt2$$