For $ a,b,c,d\geq 0 $ with $ a+b = c+d = 2 $, how to prove that $$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$$
Inequality:$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$
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2For starters, if $x,y\geq 0$, then $$(x+y)^2\geq x^2+y^2$$ – 2012-08-24
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0@PeterTamaroff if you were thinking about changing all of the expressions above like you described, that wouldn't work since after taking the square root of both sides, and making $a=2,b=0,c=1,$and $d=1$, you get $9\le5$ – 2012-08-24
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0@Sidd I don't understand what you mean. I just hinted a very obvious inequality. I don't know if it will work to prove this, and I actually scanned too fast for I thought there was an $a^2+b^2$ up there. – 2012-08-24
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0Peter's inequality looks familiar. Is there a name for that inequality – 2012-08-24
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1Just using $x\cdot y\geq0$ if $x,y\geq 0$. Expand the binomial. Obvious inequality is obvious =). – 2012-08-24
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0@PeterTamaroff. I am not askig how to get the inequality from the result, but the other way around. Did you build that from scratch? It resembles alot like the triangle inequality – 2012-08-24
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0The triangle inequality for the absolute value is $|x+y|\leq |x|+|y|$, and in general, for norms, it is $$\left\| {x+y} \right\|\leq \left\| {x} \right\|+\left\| {y} \right\|$$ Again, the above inequality is just a restatement if $x\cdot y \geq 0$ if $x,y\geq 0$. Just multiply by $2$ and sum $x^2+y^2$ to both sides. – 2012-08-24
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0I posted a [related question](http://math.stackexchange.com/questions/186557/bounding-the-product-of-a-quadrilaterals-side-lengths-in-terms-of-the-lengths-of). – 2012-08-25
1 Answers
We will make repeated use of the identity $(p^2+q^2)(r^2+s^2)=(pr-qs)^2+(ps+qr)^2$.
Applying it to the first two terms yields $$ (a^2+c^2)(a^2+d^2)=(a^2-cd)^2+(a(c+d))^2=(a^2-cd)^2+(2a)^2 \, . $$ Similarly, the third and fourth terms multiply to $(b^2-cd)^2+(2b)^2$.
Applying the identity again to these two quantities, we have $$ \left[(a^2-cd)^2+(2a)^2\right]\left[(b^2-cd)^2+(2b)^2\right] \\ = \left[(a^2-cd)(b^2-cd)-4ab\right]^2+\left[(a^2-cd)(2b)+(b^2-cd)(2a)\right]^2 \\ =: Q^2+R^2 \, , $$ and we'll now work on simplifying $Q$ and $R$ separately.
First, \begin{eqnarray} Q &=& (a^2-cd)(b^2-cd)-4ab \\ &=& a^2b^2-(a^2+b^2)cd+c^2d^2-4ab \\ &=& a^2b^2-(a^2+2ab+b^2)cd+2abcd + c^2d^2-4ab \\ &=& a^2b^2-4cd+2abcd+c^2d^2-4ab \\ &=& ab(ab+cd-4)+cd(ab+cd-4)\\ &=& (ab+cd)(ab+cd-4) \, . \end{eqnarray}
Next, \begin{eqnarray} R &=& (a^2-cd)(2b)+(b^2-cd)(2a) \\ &=& 2ba^2-2bcd+2ab^2-2acd \\ &=& 2ab(a+b)-2cd(a+b)\\ &=&4(ab-cd) \, . \end{eqnarray}
So the quantity we're trying to bound is equal to $$ \left[(ab+cd)(ab+cd-4)\right]^2+[4(ab-cd)]^2 \, . $$ Let $x=ab$, $y=cd$. Then the constraints on $a,b,c,d$ imply that $x,y$ each vary freely over $[0,1]$. So, in order to prove the inequality, it suffices to show that $$ f(x,y)=(x+y)^2(x+y-4)^2+16(x-y)^2 $$ is bounded above by 25 on the unit square.
To show this, we first make the temporary change of variables $s=x+y$, $t=x-y$; let $F(s,t)=s^2(s-4)^2+16t^2$ be the pullback of $f$ under this coordinate change. Then $F_s=2s(s-4)(2s-4)$ is nonvanishing for $0
Since $f$ is symmetric in its arguments, it suffices to examine the functions $f_0(y)=f(0,y)$ and $f_1(x)=f(x,1)$. We have $$f_0(y)=y^2(y-4)^2+16y^2=y^4-8y^3+32y^2 \\ f_0'(y)=4y^3-24y^2+64y=4y(y^2-6y+16)=4y((y-3)^2+7) \, ; $$ thus $f_0'$ is nonzero on $(0,1)$, so $f_0$ has no critical points on $(0,1)$. Also, $$ f_1(x)=(x+1)^2(x-3)^2+16(x-1)^2=((x-1)^2+1)^2\\ f_1'(x)=4(x-1)((x-2)^2+1)^2 \, ; $$ thus $f_1'$ is also nonzero on $(0,1)$, and $f_1$ has no relevant critical points. So $f$ is maximized at one of the corners of the square.
Finally, $$ f(0,0)=16 \\ f(0,1)=25 \\ f(1,1)=16 $$ and so the maximum value of $f$ is 25, as desired.
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2Supplement: Here's a little applet I made for fun to illustrate a possible geometric interpretation of the inequality. Hope it works! http://www.geogebratube.org/student/m16177. One should be able to see the product of the areas of the 4 squares displayed at the bottom. – 2012-08-24
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0@Mike: Neat! I wonder if there's a geometric proof that's less ugly than mine. You could simplify things a little by trying to show that the product of the four side lengths was at most 5; unfortunately I don't know of any geometric interpretation for the product of four sides of a quadrilateral... – 2012-08-24