Cho a,b,c>0 và abc=8. Chứng minh \(\dfrac{a-2}{a+1}+\dfrac{b-2}{b+1}+\dfrac{c-2}{c+1}\ge0\)
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Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
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Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) khi đó thu được \(xyz=1\)
Ta có:
\(\dfrac{1}{a^2\left(b+c\right)}=\dfrac{x^2}{\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{x^2yz}{y+z}=\dfrac{x}{y+z}\)
BĐT cần chứng minh được viết lại thành:\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{y}{z+x}+1\right)+\left(\dfrac{z}{x+y}+1\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)\ge\dfrac{9}{2}\)
Đánh giá cuối cùng đúng theo BĐT Cauchy
Vậy BĐT được chứng minh. Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.
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1.
Ta có:
\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)
Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:
\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)
\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2.
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
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Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)
Chứng minh rằng nếu a,b,c \(\ge\)0 và abc=1 thì
\(\dfrac{1}{2+a}+\dfrac{1}{2+b}+\dfrac{1}{2+c}\le1\)
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\(\Leftrightarrow\dfrac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\le1\)
\(\Leftrightarrow\dfrac{ab+bc+ca+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\le1\)
\(\Leftrightarrow ab+bc+ca+12\le2\left(ab+bc+ca\right)+9\)
\(\Leftrightarrow ab+bc+ca\ge3\)
Hiển nhiên đúng do: \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}=3\)
Vì abc=1 , ta đặt \(a=\dfrac{x}{y};b=\dfrac{y}{z};c=\dfrac{z}{x}\)
Điều phải chứng minh tương đương với:
\(\dfrac{1}{2+\dfrac{x}{y}}+\dfrac{1}{2+\dfrac{y}{z}}+\dfrac{1}{2+\dfrac{z}{x}}\le1\\ \Leftrightarrow\dfrac{y}{2y+x}+\dfrac{z}{2z+y}+\dfrac{x}{2x+z}\le1\\ \Leftrightarrow\dfrac{2y}{2y+x}+\dfrac{2z}{2z+y}+\dfrac{2x}{2x+z}\le2\\ \Leftrightarrow\dfrac{x}{2y+x}+\dfrac{y}{2z+y}+\dfrac{z}{2x+z}\ge1\left(1\right)\)
Áp dụng bất đẳng thức bunhiacopxki dạng phân thức ta có:
\(\dfrac{x}{2y+x}+\dfrac{y}{2z+x}+\dfrac{z}{2x+z}=\dfrac{x^2}{x^2+2xy}+\dfrac{y^2}{y^2+2zx}+\dfrac{z^2}{z^2+2xy}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
=> bài toán được chứng minh
Dấu bằng xảy ra khi x=y=z=1 <=>a=b=c=1
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2.
Từ giả thiết, ta có :
\(\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}+1-\frac{1}{1+d}\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Tương tự, ta cũng có :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{c.d.a}{\left(1+c\right)\left(1+d\right)\left(1+a\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế theo vế 4 BĐT vừa chững minh rồi rút gọn ta được :
\(abcd\le\frac{1}{81}\left(đpcm\right)\)
2) Từ \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3.\)
\(\Rightarrow\frac{1}{1+a}\ge\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}.\)(BĐT AM-GM)
Tương tự :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}.\)
Từ đó suy ra:
\(\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}\ge3.3.3.3\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^3}}\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge\frac{81abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}.\)
\(\Leftrightarrow81abcd\le1\Leftrightarrow abcd\le\frac{1}{81}\)
Dấu '=' xảy ra khi \(a=b=c=d=\frac{1}{3}.\)
3)Ta có: \(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left(a+b+2\sqrt{ab}\right)^4.\)(1)
Với \(a,b\ge0\),áp dụng BĐT AM-GM cho (a+b) và (\(2\sqrt{ab}\)) ta được
\(\left(a+b\right)+2\sqrt{ab}\ge2\sqrt{\left(a+b\right)2\sqrt{ab}}\)(2)
Từ (1) và (2) suy ra:
\(\left(\sqrt{a}+\sqrt{b}\right)^8\ge\left(2\sqrt{\left(a+b\right)2\sqrt{ab}}\right)^4\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2.\)
Dấu '=' xảy ra khi \(a+b=2\sqrt{ab}\Leftrightarrow a=b\)
1) Với \(x\le\frac{2}{3}\Rightarrow2-3x\ge0\)
Khi đó ,áp dụng bất đẳng thức AM-GM cho 2 số ta được:
\(\left(2-3x\right)+\frac{9}{2-3x}\ge2\sqrt{\left(2-3x\right)\frac{9}{2-3x}}=2.3=6\)
\(\Leftrightarrow2+\left(2-3x\right)+\frac{9}{2-3x}\ge2+6\)
\(\Leftrightarrow4-3x+\frac{9}{2-3x}\ge8\)
Dấu '=' xảy ra khi \(2-3x=\frac{9}{2-3x}\Leftrightarrow\left(2-3x\right)^2=9\Leftrightarrow2-3x=3\Leftrightarrow x=-\frac{1}{3}\)( vì 2-3x>0)
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\(\sum\dfrac{a}{\left(a^2+1\right)+2b+2}\le\sum\dfrac{a}{2\left(a+b+1\right)}=\dfrac{1}{2}\)
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Đặt \(\left\{{}\begin{matrix}x=a+b+c\\y=ab+bc+ca\end{matrix}\right.\) khi đó \(BDT\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}\le\dfrac{12+4x+y}{9+4x+2y}\)
\(\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}-1\le\dfrac{12+4x+y}{9+4x+2y}-1\)
\(\Leftrightarrow\dfrac{2x+3-xy}{x^2+2x+y+xy}\le\dfrac{3-y}{9+4x+2y}\)
\(\Leftrightarrow\dfrac{5x^2-3x^2y-xy^2-6xy+24x+y^2+3y+27}{\left(4x+2y+9\right)\left(x^2+xy+2x+y\right)}\le0\)
Đúng vì \(\dfrac{5}{3}x^2y\ge5x^2;\dfrac{x^2y}{3}\ge y^2;xy^2\ge9x;5xy\ge15x;xy\ge3y;x^2y\ge27\)