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Áp dụng bất đẳng thức Cauchy–Schwarz dạng Engel ta có :
\(VT\ge\frac{\left(2b+3c+2c+3a+2a+3b\right)^2}{a+b+c}\)
\(=\frac{\left(5a+5b+5c\right)^2}{a+b+c}=\frac{\left[5\left(a+b+c\right)\right]^2}{a+b+c}\)
\(=\frac{25\left(a+b+c\right)^2}{a+b+c}=25\left(a+b+c\right)=VP\)
=> đpcm
Đẳng thức xảy ra <=> a = b = c
VP = \(\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)
\(=\left(a-b\right).\dfrac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\left(b-c\right).\dfrac{\left(b+a\right)-\left(c+a\right)}{\left(b+a\right)\left(c+a\right)}+\left(c-b\right).\dfrac{\left(c+b\right)-\left(a+b\right)}{\left(c+b\right)\left(a+b\right)}\)
\(=\left(a-b\right).\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{b+a}\right)+\left(c-a\right).\left(\dfrac{1}{a+b}-\dfrac{1}{c+b}\right)\)
\(=\left(a-b\right).\dfrac{1}{b+c}-\left(a-b\right).\dfrac{1}{a+c}+\left(b-c\right).\dfrac{1}{c+a}-\left(b-c\right).\dfrac{1}{b+a}+\left(c-a\right).\dfrac{1}{a+b}-\left(c-a\right).\dfrac{1}{c+b}\)
\(=\left(2a-b-c\right).\dfrac{1}{b+c}+\left(2b-c-a\right).\dfrac{1}{c+a}+\left(2c-a-b\right).\dfrac{1}{a+b}\)
\(=\dfrac{2a}{b+c}-\left(b+c\right).\dfrac{1}{b+c}+\dfrac{2b}{c+a}-\left(c+a\right).\dfrac{1}{c+a}+\dfrac{2c}{a+b}-\left(a+b\right).\dfrac{1}{a+b}\)
\(=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3\left(đpcm\right)\)
\(VT=\dfrac{2a^3-a^2b-a^2c-ab^2-ac^2+2b^3-b^2c-bc^2+2c^3}{(a+b)(b+c)(c+a)} \)
\(\\=\dfrac{a^3+a^2b-2a^2b-2ab^2+ab^2+b^3+b^3+b^2c-2b^2c-2bc^2+bc^2+c^3+c^3+c^2a-2c^a+2ca^2-ca^2+a^3}{(a+b)(b+c)(c+a)}\)
\(\\=\dfrac{(a-b)^2(a+b)+(b-c)^2(b+c)+(c-a)^2(c+a)}{(a+b)(b+c)(c+a)}\)
\(\\\Rightarrow VT=\dfrac{(a-b)^2}{(c+a)(b+c)}+\dfrac{(b-c)^2}{(c+a)(a+b)}+\dfrac{(c-a)^2}{(a+b)(b+c)}=VP\)
a) Ta có: \(\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2bc-2ab-2ac+a^2+b^2+c^2-2ab-2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ca\)
\(=a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2\)
\(=4a^2+4b^2+4c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
b) Đặt x = b + c - a
y = c + a - b
z = a + b - c
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Rightarrow a+b+c=x+y+z\)
Ta có: \(\left(a+b+c\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left[xy+\left(x+y\right)z+z^2\right]\)
\(=3\left(x+y\right)\left[z^2+xy+xz+yz\right]\)
\(=3\left(x+y\right)\left[z\left(x+y\right)+y\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=3.2a.2b.2c\)
\(=24abc\) (đpcm)
\(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2=\)\(b\left(a-c\right)\left(a+c-b\right)^2\)
\(\Leftrightarrow\)\(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2-b\left(a-c\right)\left(a+c-b\right)^2=0\)
Đặt:
\(\begin{cases}a+b-c=x\\b+c-a=y\\a+c-b=z\end{cases}\)\(\hept{\Leftrightarrow\begin{cases}a=\frac{x+z}{2}\\b=\frac{x+y}{2}\\c=\frac{y+z}{2}\end{cases}}\)
\(\Leftrightarrow\)\(\frac{x+z}{2}\left(\frac{x+y}{2}-\frac{y+z}{2}\right)y^2+\frac{y+z}{2}\left(\frac{x+z}{2}-\frac{x+y}{2}\right)x^2-\frac{x+y}{2}\left(\frac{x+z}{2}-\frac{y+z}{2}\right)z^2=0\)
\(\Leftrightarrow\frac{x+z}{2}\times\frac{x-z}{2}\times y^2+\frac{z+y}{2}\times\frac{z-y}{2}\times x^2-\frac{x+y}{2}\times\frac{x-y}{2}\times z^2=0\)
\(\Leftrightarrow\frac{1}{4}\left(x+z\right)\left(x-z\right)y^2+\frac{1}{4}\left(z+y\right)\left(z-y\right)x^2-\frac{1}{4}\left(x+y\right)\left(x-y\right)z^2=0\)
\(\Leftrightarrow\frac{1}{4}\left[\left(x^2-z^2\right)y^2+\left(z^2-y^2\right)x^2\right]-\frac{1}{4}\left(x^2-y^2\right)z^2=0\)
\(\Leftrightarrow\frac{1}{4}\left(x^2y^2-z^2y^2+x^2z^2-x^2y^2\right)-\frac{1}{4}\left(x^2-y^2\right)z^2=0\)
\(\Leftrightarrow\frac{1}{4}\left(x^2-y^2\right)z^2-\frac{1}{4}\left(x^2-y^2\right)z^2=0\)
Vậy \(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2=\)\(b\left(a-c\right)\left(a+c-b\right)^2\)
a= 6 , b= 8, c = 7
CM ma bn oi