Quay lại đáp án

ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c$ là số nhỏ nhất, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$

ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c= \min\left{ a{,}b,c \right}$, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$≥4(a−c)(b−c)≥4ab+bc+ca

ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c=min\left{ a,b,c \right}$, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$

Ta có (a−b)(b−c)(c−a)≠0" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">(a−b)(b−c)(c−a)≠0(a−b)(b−c)(c−a)≠0 nên 0=(a−b)+(b−c)+(c−a)=(a−b)+(b−c)+(c−a)(a−b)(b−c)(c−a)=∑1(a−b)(b−c)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">0=(a−b)+(b−c)+(c−a)=(a−b)+(b−c)+(c−a)(a−b)(b−c)(c−a)=∑1(a−b)(b−c)0=(a−b)+(b−c)+(c−a)=(a−b)+(b−c)+(c−a)(a−b)(b−c)(c−a)=∑1(a−b)(b−c)Do đó H=∑1(a−b)2=∑1(a−b)2+2∑1(a−b)(b−c)=(∑1a−b)2=(1a−b+a−b(a−c)(b−c))2=1(a−b)2+(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 33.898em; min-height: 0px; padding-top: 1px; padding-bottom: 1px; width: 745px; position: relative;">H=∑1(a−b)2=∑1(a−b)2+2∑1(a−b)(b−c)=(∑1a−b)2=(1a−b+a−b(a−c)(b−c))2=1(a−b)2+(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)H=∑1(a−b)2=∑1(a−b)2+2∑1(a−b)(b−c)=(∑1a−b)2=(1a−b+a−b(a−c)(b−c))2=1(a−b)2+(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)Không mất tính tổng quát, giả sử c=min{a,b,c}" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">c=min{a,b,c}c=min{a,b,c}, khi đó ab+bc+ca⩾(a−c)(b−c)⩾0" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">ab+bc+ca⩾(a−c)(b−c)⩾0ab+bc+ca⩾(a−c)(b−c)⩾0 nên:H⩾21(a−b)2.(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)⩾4(a−c)(b−c)⩾4ab+bc+ca=1" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">H⩾2√1(a−b)2.(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)⩾4(a−c)(b−c)⩾4ab+bc+ca