Cho $a,b,c>0$. [Bài toán]: CMR: $\displaystyle{\frac{a}{a+b+7c}+\frac{b}{b+c+7a}+\frac{c}{c+a+7b}+\frac{2}{3}.\frac{ab+bc+ca}{a^2+b^2+c^2}\leq 1}$ Lời Giải BĐT cần CM tương đương: $\sum \displaystyle{\frac{a(a+b+c)}{a+b+7c}+\frac{2}{3}.\frac{(ab+bc+ca)(a+b+c)}{a^2+b^2+c^2}\leq a+b+c}$ $\Leftrightarrow \sum \displaystyle{\left ( a-\frac{6ac}{a+b+7c} \right )+\frac{2}{3}.\frac{(ab+bc+ca)(a+b+c)}{a^2+b^2+c^2}}\leq a+b+c$ $\Leftrightarrow \sum \displaystyle{\frac{ac}{a+b+7c}\geq \frac{1}{9}.\frac{(ab+bc+ca)(a+b+c)}{a^2+b^2+c^2}}$ Áp dụng BĐT $Cauchy-Schwarz$: Ta có: $\sum \displaystyle{\frac{ac}{a+b+7c}=\sum \frac{(ab)^2}{ac(a+b+7c)}\geq \frac{(ab+bc+ca)^2}{\sum ac(a+7c)+3abc}}$ Ta phải CM: $\displaystyle{\frac{(ab+bc+ca)^2}{\sum ac(a+7c)+3abc}\geq \frac{1}{9}.\frac{(ab+bc+ca)(a+b+c)}{a^2+b^2+c^2}}$ $\Leftrightarrow 9(ab+bc+ca)(a^2+b^2+c^2)\geq (a+b+c)[\Sigma ac(a+7c)+3abc]$ $\Leftrightarrow 9[\Sigma ab(a^2+b^2)]+9abc(a+b+c)\geq 3abc(a+b+c)+\sum (a^3c+a^2bc+8a^2b^2+7abc^2+7ac^3)$ $\Leftrightarrow 8(\Sigma a^3b)+2(\Sigma ab^3)\geq 2abc(a+b+c)+8(\Sigma a^2b^2)$ $\Leftrightarrow \sum (4a^3b-4a^2b^2+ab^3)\geq abc(a+b+c)$ $\Leftrightarrow \sum ab(2a-b)^2\geq abc(a+b+c)\qquad (1)$ Áp dụng BĐT $Cauchy-Schwarz$: $[\Sigma ab(2a-b)^2]\left ( \Sigma \displaystyle{\frac{1}{ab}} \right )\geq (2a-b+2b-c+2c-a)^2\Leftrightarrow (1)$ Vậy BĐT được chứng minh. Dấu bằng xảy ra $\Leftrightarrow a=b=c$ $\blacksquare$

Inequality 15

Cho $a, b, c > 0$ .
[Bài toán]: CMR: $\frac{a}{a + b + 7 c} + \frac{b}{b + c + 7 a} + \frac{c}{c + a + 7 b} + \frac{2}{3} . \frac{ab + b c + c a}{a ^{2} + b ^{2} + c ^{2}} \leq 1$
Lời Giải
BĐT cần CM tương đương:
$\sum \frac{a ( a + b + c )}{a + b + 7 c} + \frac{2}{3} . \frac{( ab + b c + c a ) ( a + b + c )}{a ^{2} + b ^{2} + c ^{2}} \leq a + b + c$
$\Leftrightarrow \sum (a - \frac{6 a c}{a + b + 7 c}) + \frac{2}{3} . \frac{( ab + b c + c a ) ( a + b + c )}{a ^{2} + b ^{2} + c ^{2}} \leq a + b + c$
$\Leftrightarrow \sum \frac{a c}{a + b + 7 c} \geq \frac{1}{9} . \frac{( ab + b c + c a ) ( a + b + c )}{a ^{2} + b ^{2} + c ^{2}}$
Áp dụng BĐT $C a u c h y - S c h w a r z$ :
Ta có: $\sum \frac{a c}{a + b + 7 c} = \sum \frac{( ab ) ^{2}}{a c ( a + b + 7 c )} \geq \frac{( ab + b c + c a ) ^{2}}{\sum a c ( a + 7 c ) + 3 ab c}$
Ta phải CM: $\frac{( ab + b c + c a ) ^{2}}{\sum a c ( a + 7 c ) + 3 ab c} \geq \frac{1}{9} . \frac{( ab + b c + c a ) ( a + b + c )}{a ^{2} + b ^{2} + c ^{2}}$
$\Leftrightarrow 9 (ab + b c + c a) (a^{2} + b^{2} + c^{2}) \geq (a + b + c) [Σ a c (a + 7 c) + 3 ab c]$
$\Leftrightarrow 9 [Σ ab (a^{2} + b^{2})] + 9 ab c (a + b + c) \geq 3 ab c (a + b + c) + \sum (a^{3} c + a^{2} b c + 8 a^{2} b^{2} + 7 ab c^{2} + 7 a c^{3})$
$\Leftrightarrow 8 (Σ a^{3} b) + 2 (Σ a b^{3}) \geq 2 ab c (a + b + c) + 8 (Σ a^{2} b^{2})$
$\Leftrightarrow \sum (4 a^{3} b - 4 a^{2} b^{2} + a b^{3}) \geq ab c (a + b + c)$
$\Leftrightarrow \sum ab (2 a - b)^{2} \geq ab c (a + b + c) (1)$
Áp dụng BĐT $C a u c h y - S c h w a r z$ :
$[Σ ab (2 a - b)^{2}] (Σ \frac{1}{ab}) \geq (2 a - b + 2 b - c + 2 c - a)^{2} \Leftrightarrow (1)$
Vậy BĐT được chứng minh. Dấu bằng xảy ra $\Leftrightarrow a = b = c$ $■$

Inequality 15

Đề bài

Mô tả