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Algebra Difficulty 7.7 National olympiad, round 2 Prove it

31. Let aia_{i} be positive real numbers (i=1,2,⋯ ,n)(i=1,2, \cdots, n), and let bk=a1+a2+⋯+akk(k=1,2,⋯ ,n)b_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n), Cn=(a1−b1)2+(a2−b2)2+⋯+(an−bn)2C_{n}=\left(a_{1}-b_{1}\right)^{2}+\left(a_{2}-b_{2}\right)^{2}+\cdots+\left(a_{n}-b_{n}\right)^{2}, Dn=(a1−bn)2+(a2−bn)2+⋯+(an−bn)2D_{n}=\left(a_{1}-b_{n}\right)^{2}+\left(a_{2}-b_{n}\right)^{2}+\cdots+\left(a_{n}-b_{n}\right)^{2}. Prove that: Cn⩽Dn⩽2CnC_{n} \leqslant D_{n} \leqslant 2 C_{n}. (1978 All-Soviet Union Mathematical Olympiad Problem)

Solution

31. Construct the sequences xn=2Cn−Dn,yn=Dn−Cn,n∈N∗x_{n}=2 C_{n}-D_{n}, y_{n}=D_{n}-C_{n}, n \in \mathbf{N}^{*}, then
xn+1−xn=2(Cn+1−Cn)−(Dn+1−Dn)=2(an+1−bn+1)2−(an+1−bn+1)2−n(bn+12−bn2)+2(bn+1−bn)(a1+a2+⋯+an)=(an+1−bn+1)2−n(bn+12−bn2)+2nbn(bn+1−bn)=[(n+1)bn+1−nbn−bn+1]2−n(bn+12−bn2)+2n(bnbn+1−bn2)=(n2−n)(bn+1−bn)2⩾0\begin{array}{l} x_{n+1}-x_{n}=2\left(C_{n+1}-C_{n}\right)-\left(D_{n+1}-D_{n}\right)= \\ 2\left(a_{n+1}-b_{n+1}\right)^{2}-\left(a_{n+1}-b_{n+1}\right)^{2}- \\ n\left(b_{n+1}^{2}-b_{n}^{2}\right)+2\left(b_{n+1}-b_{n}\right)\left(a_{1}+a_{2}+\cdots+a_{n}\right)= \\ \left(a_{n+1}-b_{n+1}\right)^{2}-n\left(b_{n+1}^{2}-b_{n}^{2}\right)+2 n b_{n}\left(b_{n+1}-b_{n}\right)= \\ {\left[(n+1) b_{n+1}-n b_{n}-b_{n+1}\right]^{2}-n\left(b_{n+1}^{2}-b_{n}^{2}\right)+} \\ 2 n\left(b_{n} b_{n+1}-b_{n}^{2}\right)=\left(n^{2}-n\right)\left(b_{n+1}-b_{n}\right)^{2} \geqslant 0 \end{array}

Also, x1=2C1−D1=(a1−b1)2⩾0x_{1}=2 C_{1}-D_{1}=\left(a_{1}-b_{1}\right)^{2} \geqslant 0, hence for all n∈N∗,xn⩾0n \in \mathbf{N}^{*}, x_{n} \geqslant 0.
Similarly,
yn+1−yn=n(bn+12−bn2)−2(bn+1−bn)(a1+a2+⋯+an)=n(bn+1−bn)2⩾0\begin{aligned} y_{n+1}-y_{n}= & n\left(b_{n+1}^{2}-b_{n}^{2}\right)-2\left(b_{n+1}-b_{n}\right)\left(a_{1}+a_{2}+\cdots+a_{n}\right)= \\ & n\left(b_{n+1}-b_{n}\right)^{2} \geqslant 0 \end{aligned}

Also, y1=D1−C1=0y_{1}=D_{1}-C_{1}=0, hence for all n∈N∗,yn⩾0n \in \mathbf{N}^{*}, y_{n} \geqslant 0.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.