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SCó :\(A=\) \(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\)=\(\sqrt{\frac{1}{1+\left(\frac{b+c}{a}\right)^3}}\) ( chia tử mẫu cho \(a^3\) ) =\(\frac{1}{\sqrt{1+\left(\frac{b+c}{a}\right)^3}}\)
Lại có\(\sqrt{1+\left(\frac{b+c}{a}\right)^3}\) =\(\sqrt{\left(1+\frac{b+c}{a}\right)\left[1-\frac{b+c}{a}+\left(\frac{b+c}{a}\right)^2\right]}\)( hằng đẳng thức )
\(\le\)\(\frac{2+\left(\frac{b+c}{a}\right)^2}{2}\)( áp dụng \(\sqrt{xy}\le\frac{x+y}{2}\))
Nên \(\frac{1}{\sqrt{1+\left(\frac{b+c}{a}\right)^3}}\ge\frac{2}{2+\left(\frac{b+c}{a}\right)^2}\)\(=\frac{2a^2}{2a^2+\left(b+c\right)^2}\ge\frac{2a^2}{2a^2+2b^2+2c^2}\)( vì \(\left(b+c\right)^2\le2b^2+2c^2\)) . TỪ ĐÓ SUY RA :
\(A=\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}\ge\frac{a^2}{a^2+b^2+c^2}\)
Cmtt có : \(B=\sqrt[]{\frac{b^3}{b^3+\left(c+a\right)^3}}\ge\frac{b^2}{a^2+b^2+c^2}\)
\(C=\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge\frac{c^2}{a^2+b^2+c^2}\)
VẬY \(A+B+C\ge1\)
a) 9x2 - 36
=(3x)2-62
=(3x-6)(3x+6)
=4(x-3)(x+3)
b) 2x3y-4x2y2+2xy3
=2xy(x2-2xy+y2)
=2xy(x-y)2
c) ab - b2-a+b
=ab-a-b2+b
=(ab-a)-(b2-b)
=a(b-1)-b(b-1)
=(b-1)(a-b)
P/s đùng để ý đến câu trả lời của mình
Không làm mất tính tổng quát của bài toán, giả sử \(a\ge b\ge c\)(1)
Có \(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}=\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
Từ (1) => \(\hept{\begin{cases}\frac{2}{a}\le\frac{1}{a}+\frac{1}{b}\\\frac{2}{b}\le\frac{1}{b}+\frac{1}{c}\\\frac{2}{c}\le\frac{1}{a}+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{\frac{2}{a}}\le\sqrt{\frac{1}{a}+\frac{1}{b}}\\\sqrt{\frac{2}{b}}\le\sqrt{\frac{1}{b}+\frac{1}{c}}\\\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{a}+\frac{1}{c}}\end{cases}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)
Ta có đpcm
\(P=\frac{2a}{2\sqrt{\left(b+1\right)\left(b^2-b+1\right)}+2}+\frac{2b}{2\sqrt{\left(c+1\right)\left(c^2-c+1\right)}+2}\)\(+\frac{2c}{2\sqrt{\left(a+1\right)\left(a^2-a+1\right)}+2}\)
\(P\ge\frac{2a}{b^2+4}+\frac{2b}{c^2+4}+\frac{2c}{a^2+4}\)
\(2P\ge\frac{4a}{b^2+4}+\frac{4b}{c^2+4}+\frac{4c}{a^2+4}=a-\frac{ab^2}{b^2+4}+b-\frac{bc^2}{c^2+4}+a-\frac{ca^2}{a^2+4}\)
\(2p\ge a+b+c-\left(\frac{ab^2}{4b}+\frac{bc^2}{4c}+\frac{ca^2}{4a}\right)\)
\(2P\ge6-\frac{1}{4}\left(ab+bc+ca\right)\ge6-\frac{1}{12}\left(a+b+c\right)^2=3\)
\(\Rightarrow P\ge\frac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c=2\)
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sai đề nhé ở đây, min nó là 16 mà 6 căn 6=14 thôi, mà cái điểm rơi cũng ngộ nữa :))
Nếu bạn đã nói sai thì cho mình giải thử nhé!
Áp dụng BĐT Bunhiacopxky - Cauchy - Schwarz, ta có:
\(\left(ax+by+cz\right)^2\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)\(\Rightarrow\sqrt{a^2+b^2+c^2}\cdot\sqrt{x^2+y^2+z^2}\ge ax+by+cz\)(với a, b, c, x, y, z là những số dương)
\(\Rightarrow\sqrt{2+18+4}\cdot\sqrt{\frac{8}{a^2}+\frac{9b^2}{2}+\frac{c^2a^2}{4}}\ge\sqrt{2}\cdot\frac{2\sqrt{2}}{a}+3\sqrt{2}\cdot\frac{3b}{\sqrt{2}}+2\cdot\frac{ca}{2}\)
\(\Leftrightarrow\sqrt{24}\cdot\sqrt{\frac{8}{a^2}+\frac{9b^2}{2}+\frac{c^2a^2}{4}}\ge\frac{4}{a}+9b+ca\)(1)
Tương tự ta có: \(\sqrt{24}.\sqrt{\frac{8}{b^2}+\frac{9c^2}{2}+\frac{a^2b^2}{4}}\ge\frac{4}{b}+9c+ab\)(2)
\(\sqrt{24}\cdot\sqrt{\frac{8}{c^2}+\frac{9a^2}{2}+\frac{b^2c^2}{4}}\ge\frac{4}{c}+9a+bc\)(3)
Cộng vế theo vế (1), (2) và (3) ta được: \(\sqrt{24}\cdot\left(VT\right)\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}+9\left(a+b+c\right)+ab+bc+ca\)
\(=\left(\frac{4}{a}+a\right)+\left(\frac{4}{b}+b\right)+\left(\frac{4}{c}+c\right)+\left(2a+bc\right)+\left(2b+ca\right)+\left(2c+ab\right)\)\(+6\left(a+b+c\right)\)\(\ge2\sqrt{\frac{4}{a}\cdot a}+2\sqrt{\frac{4}{b}\cdot b}+2\sqrt{\frac{4}{c}\cdot c}+2\sqrt{2abc}+2\sqrt{2abc}+2\sqrt{2abc}\)\(+6\left(a+b+c\right)\)\(=12+6\left(a+b+c+\sqrt{2abc}\right)\ge12+6\cdot10=72\)
\(\Rightarrow VT\ge\frac{72}{\sqrt{24}}=6\sqrt{6}\)
Dấu ''='' xảy ra khi: \(\hept{\begin{cases}a+b+c+\sqrt{2abc}=10\\VT=6\sqrt{6}\end{cases}\Leftrightarrow a=b=c=2}\)
Vậy ta được ĐPCM
Đặt \(\left\{{}\begin{matrix}b+c-a=x>0\\c+a-b=y>0\\a+b-c=z>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{y+z}{2}\\b=\frac{z+x}{2}\\c=\frac{x+y}{2}\end{matrix}\right.\)
BĐT trở thành: \(\frac{\sqrt{y+z}}{\sqrt{2}x}+\frac{\sqrt{z+x}}{\sqrt{2}y}+\frac{\sqrt{x+y}}{\sqrt{2}z}\ge\frac{x+y+z}{\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8}}}\)
\(\Leftrightarrow\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\ge\frac{4\left(x+y+z\right)}{\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}}\)
\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}+\frac{\left(z+x\right)\sqrt{\left(y+z\right)\left(y+x\right)}}{y}+\frac{\left(x+y\right)\sqrt{\left(z+x\right)\left(z+y\right)}}{z}\ge4\left(x+y+z\right)\)
Ta có:
\(\frac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge\frac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}=y+z+\frac{\left(y+z\right)\sqrt{yz}}{x}\ge y+z+\frac{2yz}{x}\)
Tương tự: \(\frac{\left(z+x\right)\sqrt{\left(y+z\right)\left(y+x\right)}}{y}\ge z+x+\frac{2zx}{y}\) ; \(\frac{\left(x+y\right)\sqrt{\left(z+x\right)\left(z+y\right)}}{z}\ge x+y+\frac{2xy}{z}\)
Cộng vế với vế:
\(VT\ge2\left(x+y+z\right)+2\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\right)\ge2\left(x+y+z\right)+2\left(x+y+z\right)\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
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