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\(P=\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}=\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}\)
vì a,b,c là 3 cạnh của 1 tam giác áp dụng bđt tam giác có:
\(\hept{\begin{cases}b+c>a\Rightarrow2b+2c>a\Rightarrow2ab+2ac>a^2\Rightarrow2ab+2ac-a^2>0\\c+a>b\Rightarrow2c+2a>b\Rightarrow2bc+2ab>b^2\Rightarrow2bc+2ab-b^2>0\\a+b>c\Rightarrow2a+2b>c\Rightarrow2ac+2bc>c^2\Rightarrow2ac+2bc-c^2>0\end{cases}}\)
\(\Rightarrow\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}>0\)áp dụng bđt cauchy schawazt dạng enge ta có:
\(\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}>=\)
\(\frac{\left(a+b+c\right)^2}{2ab+2ac-a^2+2bc+2ab-b^2+2ac+2bc-c^2}=\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-\left(a^2+b^2+c^2\right)}\left(1\right)\)
vì \(a^2+b^2+c^2>=ab+ac+bc\Rightarrow4ab+4ac+4bc-\left(a^2+b^2+c^2\right)< =\)
\(4ab+4ac+4bc-\left(ab+ac+bc\right)\)mà \(\left(a+b+c\right)^2>0\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-\left(a^2+b^2+c^2\right)}>=\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-\left(ab+ac+bc\right)}\)(2)
\(=\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-ab-ac-bc}=\frac{\left(a+b+c\right)^2}{3ab+3ac+3bc}=\frac{a^2+b^2+c^2+2ab+2ac+2bc}{3ab+3ac+3bc}\)
\(>=\frac{ab+ac+bc+2ab+2ac+2bc}{3ab+3ac+3bc}=\frac{3ab+3ac+3bc}{3ab+3ac+3bc}=1\)(3)
từ (1)(2)(3)\(\Rightarrow\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}>=1\)
\(\Rightarrow P=\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}>=1\)
dấu = xảy ra khi a=b=c
vậy min P là 1 khi a=b=c
Đặt \(\hept{\begin{cases}x=2b+2c-a\\y=2c+2a-b\\z=2a+2b-c\end{cases}}\)
Vì a,b,c là độ dài ba cạnh của 1 tam giác nên \(x,y,z>0\)
Khi đó :
\(\Rightarrow\hept{\begin{cases}a=\frac{2y+2z-x}{9}\\b=\frac{2z+2x-y}{9}\\c=\frac{2x+2y-z}{9}\end{cases}}\)
Ta có bất đẳng thức mới theo ẩn x,y,z :
\(\frac{2y+2z-x}{9x}+\frac{2z+2x-y}{9y}+\frac{2x+2y-z}{9z}\ge1\)
\(\Leftrightarrow\frac{2}{9}\left(\frac{y}{x}+\frac{z}{x}\right)+\frac{2}{9}\left(\frac{z}{y}+\frac{x}{y}\right)+\frac{2}{9}\left(\frac{x}{z}+\frac{y}{z}\right)-\frac{1}{3}\ge1\)
\(\Leftrightarrow\frac{2}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{2}{9}\left(\frac{y}{z}+\frac{z}{y}\right)+\frac{2}{9}\left(\frac{z}{x}+\frac{x}{z}\right)-\frac{1}{3}\ge1\)
Ta chứng minh bất đẳng thức phụ sau :
\(\frac{a}{b}+\frac{b}{a}\ge2\forall a,b>0\)
Thật vậy : \(\frac{a}{b}+\frac{b}{a}\ge2\)
\(\Leftrightarrow\frac{a^2}{ab}+\frac{b^2}{ab}\ge2\)
\(\Leftrightarrow\frac{a^2+b^2}{ab}-2\ge0\)
\(\Leftrightarrow\frac{a^2+b^2-2ab}{ab}\ge0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)(luôn đúng \(\forall a,b>0\))
Áp dụng , ta được :
\(\frac{2}{9}.2+\frac{2}{9}.2+\frac{2}{9}.2-\frac{1}{3}\ge1\)
\(\Leftrightarrow\frac{12}{9}-\frac{1}{3}\ge1\)
\(\Leftrightarrow\frac{9}{9}\ge1\)(đúng)
Vậy bất đẳng thức được chứng minh
P = \(\frac{a^2c}{a^2c+c^2b+b^2a+}+\frac{b^2a}{b^2a+a^2c+c^2b}+\frac{c^2b}{c^2b+b^2a+a^2c}\)
P = \(\frac{a^2c+b^2a+c^2b}{a^2c+c^2b+b^2a}=1\)
\(P=\frac{\frac{a}{b}}{\frac{a}{b}+\frac{c}{a}+\frac{b}{c}}+\frac{\frac{b}{c}}{\frac{b}{c}+\frac{a}{b}+\frac{c}{a}}+\frac{\frac{c}{a}}{\frac{c}{a}+\frac{b}{c}+\frac{a}{b}}=\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}=1\)
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
Ta có:
A = \(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{3b+2a}=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{3bc+2ac}\)
A \(\ge\frac{\left(a+b+c\right)^2}{2ab+3ac+2bc+3ab+3bc+2ac}\)(bđt svacxo \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\))
A \(\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ac\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{5\left(a+b+c\right)^2}{3}}\) (bđt \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)(*)
CM bđt * <=> \(3xy+3yz+3xz\le x^2+y^2+z^2+2xz+2xy+2yz\)
<=> \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\ge0\) (luôn đúng)
<=> A \(\ge\frac{3}{5}\) --> ĐPCM
Cần CM bĐT phụ sau : \(\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{1}{a+b}\left(1\right)\)
Có \(a+b\ge2\sqrt{ab},\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\)
\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\Rightarrow\) (1) đúng
Áp dụng (1) ta có \(\frac{1}{2a+b+c}=\frac{1}{\left(a+b+c\right)+a}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{a+b+c}\right)\left(2\right)\)
Tương tự có \(\frac{1}{a+2b+c}\le\frac{1}{4}\left(\frac{1}{a+b+c}+\frac{1}{b}\right)\left(3\right),\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a+b+c}+\frac{1}{c}\right)\left(4\right)̸\)
Cọng (2),(3) và (4) có \(VT\le\frac{1}{4}\left(\frac{3}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+c}=\frac{1}{a+a+b+c}\le\frac{1}{4}\left(\frac{1}{a+a}+\frac{1}{b+c}\right)\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Tương tự ta có: \(\frac{1}{a+2b+c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{2}{b}+\frac{1}{c}\right)\) ; \(\frac{1}{a+b+2c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\right)\)
Cộng vế với vế:
\(VT\le\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{2}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta CM BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},a+b\ge2\sqrt{ab}\)( co si với a,b>0)
Suy ra \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(a+b\right)\ge4\RightarrowĐPCM\)\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)
a/Áp dụng (1) có
\(\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\left(2\right)\).Tương tự ta cũng có:
\(\frac{1}{b+c+2a}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\left(3\right),\frac{1}{c+a+2b}\le\frac{1}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\left(4\right)\)
Cộng (2),(3) và (4) có \(VT\le\frac{1}{4}.\left(6+6\right)=3\left(ĐPCM\right)\)
b/Áp dụng (1) có:
\(\frac{1}{3a+3b+2c}=\frac{1}{\left(a+b+2c\right)+2\left(a+b\right)}\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{2\left(a+b\right)}\right)\left(5\right)\)
Tương tự có: \(\frac{1}{3a+2b+3c}\le\frac{1}{4}\left(\frac{1}{a+c+2b}+\frac{1}{2\left(a+c\right)}\right)\left(6\right)\)
\(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{2a+b+c}+\frac{1}{2\left(b+c\right)}\right)\left(7\right)\)
Cộng (5),(6) và (7) có:
\(VT\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{a+c+2b}+\frac{1}{2a+b+c}+\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\right)\le\frac{1}{4}.9=\frac{3}{2}\)
Trước hết ta chứng minh các bđt : \(a^7+b^7\ge a^2b^2\left(a^3+b^3\right)\left(1\right)\)
Thật vậy:
\(\left(1\right)\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\ge0\)(luôn đúng)
Lại có : \(a^3+b^3+1\ge ab\left(a+b+1\right)\)
\(\Leftrightarrow a^3+b^3+abc\ge ab\left(a+b+1\right)\)
mà \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+1\right)\)(luôn đúng)
Áp dụng các bđt trên vào bài toán ta có
∑\(\frac{a^2b^2}{a^7+a^2b^2+b^7}\le\)∑\(\frac{a^2b^2}{a^3b^3\left(a+b+c\right)}\le\)∑\(\frac{a+b+c}{a+b+c}=1\)
Bất đẳng thức được chứng minh
Dấu "=" xảy ra khi a=b=c=1
Em xem lại dòng thứ 4 và giải thích lại giúp cô với! ko đúng hoặc bị nhầm
Theo BĐT AM-GM \(VT\ge6\sqrt[3]{\frac{abc}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)
\(\sqrt{\left(b+c-a\right)\left(c+a-b\right)}\le\frac{1}{2}\left(b+c-a+c+a-b\right)=c\)
Tương tự ta có: \(\sqrt{\left(c+a-b\right)\left(a+b-c\right)}\le b\)
\(\sqrt{\left(b+c-a\right)\left(a+b-c\right)}\le b\)
\(\Rightarrow\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\le abc\)
\(\Leftrightarrow6\sqrt[3]{\frac{abc}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\ge6\)
\(\Rightarrow\frac{2a}{b+c-a}+\frac{2b}{a+c-b}+\frac{2c}{a+b-c}\ge6\)
Đpcm