Larger Squares

Method 1:

Let $A = 1^2 + 2^2 + 3^2 + 4^2, B^2= 5^2 + 6^2 + 7^2 + 8^2 , C = 9^2 + 10^2 + 11^2 + 12^2, D = 13^2 + 14^2 + 15^2 + 16^2$ .

Essentially we want to prove that $D+A >B+C$ , which is equivalent to $D-C > B-A$ .

Recall the difference of 2 squares identity , $a^2 - b^2 = (a+b)(a-b)$ .

So $\begin{aligned}D - C &= &(13^2 - 9^2) + (14^2 - 10^2) + (15^2 - 11^2) + (16^2 - 12^2) \\ &=& (13-9)(13+9) + (14-10)(14+10) + (15-11)(15+11) + (16-12)(16+12) \\ &=& 4(9+10+11+12+13+14+15+16) . \end{aligned}$

Likewise, we should get $B-A = 4(1+2+3+4+5+6+7+8)$ .

Since it is obvious that $9+10+11+12+13+14+15+16 > 1+2+3+4+5+6+7+8$ , then $D-C > B-A$ , and we're done.

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Method 2 : Let $f(x) = x^2 + (17-x)^2$ . We want to prove that $f(1) + f(2) + f(3) + f(4) > f(5) + f(6) + f(7) + f(8)$ , which is equivalent to $[f(1) - f(5) ] + [ f(2) - f(6) ] + [ f(3) - f(7) ] + [ f(4) - f(8) ] > 0$ .

Let $g(x) = f(x) - f(x-4) = x^2 - (x-4)^2 + (17-x)^2 - (13 - x)^2 = 8(x-2) + 8(15-x) = 8\times13 > 0$ . Thus

$[f(1) - f(5) ] + [ f(2) - f(6) ] + [ f(3) - f(7) ] + [ f(4) - f(8) ] = g(1) + g(2) + g(3) + g(4) > 0 .$

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Method 3: Let $E = 1^2 + 2^2 + \ldots + 16^2 , F = 1^2 + 2^2 + \cdots + 12^2, G = 1^2 + 2^2 + 3^2 + 4^2$ .

We want to prove that $E - 2F+ 2G > 0 \qquad \Leftrightarrow \qquad E > 2F - 2G$ .

Recall the algebraic identity , $1^2 + 2^2 + \cdots + n^2 = \dfrac16 n(n+1)(2n+1)$ , then proving the inequality above is equivalent to

$6E > 2(6F-2G) \qquad \Leftrightarrow \qquad 16(17)(33) > 2 [ 12(13)(25) - 4(5)(9) ] .$

Since it is fairly easy to verify that $16\times 17\times33 > 12 \times 25 \times26$ , the result follows.

$\color{#3D99F6} 1^2+2^2+3^3+4^2+13^2+14^2+15^2+16^2 = 876$

$\color{#D61F06} 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 = 620$

Hence $\color{#3D99F6}{876} \color{#20A900}> \color{#D61F06} {620}$