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a) Ta có: (a - 1)( b - 1)(c - 1) = abc - ab - bc - ac + a + b +c - 1 (*)
Mà abc =1 => (*) = (1 - 1) + (a + b + c) - (ab + bc + ac)
= ( a + b + c ) - ( ab + ac + bc)
\(\ge\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) - ( ab + ac +bc )
= \(\dfrac{ab+ac+bc}{abc}\) - ab - ac - bc
= ab + bc + ac - ab - ac - bc = 0 ( do abc =1)
=> đpcm
Với mọi a , b , c \(\in\)R ta luôn có :
\(a^2\)+ \(b^2\)+ \(c^2\)> hoặc = \(2bc+2ca-2ab\left(1\right)\)
Ta cần chứng minh ( 1 ) là bất đẳng thức đúng
\(\Leftrightarrow\)\(a^2\)+ \(b^2\)+ \(c^2\)+ 2ab - 2bc - 2ca > hoặc = 0
\(\Leftrightarrow\)\(\left(a+b-c\right)^2\) > hoặc = 0 ( 2 )
Bất đẳng thức ( 2 ) luôn đúng với mọi a ; b ; c mà các phép biến đổi trên tương ứng
Nên bất đẳng thức ( 1 ) được chứng minh
Xảy ra khi và chỉ khi a + b = c
Mà \(a^2\)+ \(b^2\)+ \(c^2\)= \(\frac{5}{3}\)( gt )
Mà \(\frac{5}{3}\)= \(1\frac{2}{3}\)< 2 ( 3 )
Từ ( 1 ) kết hợp với ( 3 ) ta có thể viết :
2bc + 2ca - 2ab < hoặc = \(a^2\)+ \(b^2\)+ \(c^2\)< 2
\(\Rightarrow\)2bc + 2ca - 2ab < 2
Vì a ; b ; c > 0 nên chia cả 2 vế của bđt cho 2abc
\(\frac{2bc+2ca-2ab}{2abc}< \frac{2}{2abc}\)
\(\Leftrightarrow\)\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}\)
Vậy với a ; b ; c là các số dương thỏa mãn điều kiện : \(a^2\)+ \(b^2\)+ \(c^2\)= \(\frac{5}{3}\)thì ta luôn chứng minh được :
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(a^3+a\geq 2a^2; b^3+b\geq 2b^2; c^3+c\geq 2c^2\)
\(\Rightarrow A=\frac{a}{a^3+a+1}+\frac{b}{b^3+b+1}+\frac{c}{c^3+c+1}\leq \frac{a}{2a^2+1}+\frac{b}{2b^2+1}+\frac{c}{2c^2+1}\)
\(\leq \frac{a}{a^2+2a}+\frac{b}{b^2+2b}+\frac{c}{c^2+2c}\)
hay \(A\leq \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}(1)\)
Vì $abc=1$ nên đặt \((a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})(x,y,z>0)\)
Khi đó:
\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}=\frac{1}{2}(1-\frac{x}{x+2y}+1-\frac{y}{y+2z}+1-\frac{z}{z+2x})\)
\(=\frac{3}{2}-\frac{1}{2}(\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2zy}+\frac{z^2}{z^2+2xz})\)
\(\leq \frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{x^2+2xy+y^2+2zy+z^2+2xz}=\frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{(x+y+z)^2}=1(2)\) (theo BĐT Cauchy-Schwarz)
Từ \((1);(2)\Rightarrow A\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
bai n ay la bai o dau ma dau cung thay nhi
\(\left(a^{\dfrac{1}{3}};b^{\dfrac{1}{3}};c^{\dfrac{1}{3}}\right)->\left(x;y;z>0\right)\left(xyz=1\right)\)\(\RightarrowΣ\dfrac{x^3}{x^9+x^3+1}\le1\)
\(\dfrac{x^3}{x^9+x^3+1}\le\dfrac{x^2+1}{2\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow-\dfrac{\left(x-1\right)^2\left(x^9+2x^8+4x^7+6x^6+6x^5+6x^4+5x^3+4x^2+2x+1\right)}{2\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^9+x^3+\right)}\le0\)
\(\Rightarrow VT\le\dfrac{1}{2}\cdot2=1=VP\)
a=b=c=x=y=z=1
Câu a :
Theo BĐT cauchy schwar ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}=\dfrac{9}{x+y+z}\)
\(\Rightarrow\left(x+y+z\right)\left(\dfrac{9}{x+y+z}\right)\ge9\)
Câu b : Sửa lại đề nha :
Theo BĐT cauchy schwar ta có :
\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=\dfrac{9}{\left(a+b+c\right)^2}\)
Vì \(a+b+c\le\Rightarrow\left(a+b+c\right)^2\le1\)
\(\Rightarrow\) \(\dfrac{9}{\left(a+b+c\right)^2}\ge9\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}=3+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)Ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2;\dfrac{c}{a}+\dfrac{a}{c}\ge2;\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3+2+2+2=9\)
Gọi \(A=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) Ta có:
\(A=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)\(=\left(\dfrac{a}{a}+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{b}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{c}{c}\right)\)
\(=\left(1+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+1+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+1\right)\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)
\(=3+\left(\dfrac{a^2}{ab}+\dfrac{b^2}{ab}\right)+\left(\dfrac{b^2}{bc}+\dfrac{c^2}{bc}\right)+\left(\dfrac{c^2}{ac}+\dfrac{a^2}{ac}\right)\)
\(=3+\dfrac{a^2+b^2}{ab}+\dfrac{b^2+c^2}{bc}+\dfrac{c^2+a^2}{ac}\)
\(=3+\dfrac{a^2-2ab+b^2+2ab}{ab}+\dfrac{b^2-2bc+c^2+2bc}{bc}+\dfrac{c^2-2ac+a^2+2ac}{ac}\)
\(=3+\dfrac{\left(a-b\right)^2+2ab}{ab}+\dfrac{\left(b-c\right)^2+2bc}{bc}+\dfrac{\left(c-a\right)^2+2ac}{ac}\)
\(=3+\dfrac{\left(a-b\right)^2}{ab}+2+\dfrac{\left(b-c\right)^2}{bc}+2+\dfrac{\left(c-a\right)^2}{ac}+2\)
\(=9+\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ac}\)
Ta thấy: \(\dfrac{\left(a-b\right)^2}{ab}\ge0\) với \(\forall\) a, b
\(\dfrac{\left(b-c\right)^2}{bc}\ge0\) với \(\forall\) b, c
\(\dfrac{\left(c-a\right)^2}{ac}\ge\) 0 với \(\forall\) a, c
=> \(A\ge9\).
Vậy...
Chúc bạn học tốt!
dạng này chắc chắc là phải dùng AM-GM ngược dấu rồi :)
Ta có:
\(\dfrac{1+b}{1+4a^2}=1+b-\dfrac{4a^2\left(b+1\right)}{4a^2+1}\ge1+b-\dfrac{4a^2\left(b+1\right)}{4a}=1+b-a\left(b+1\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1+c}{1+4b^2}\ge1+c-b\left(c+1\right);\dfrac{1+a}{1+4c^2}\ge1+a-c\left(a+1\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+c^2}\)
\(\ge3+\left(a+b+c\right)-\left(ab+bc+ca\right)-\left(a+b+c\right)\)
\(=3-\dfrac{1}{3}\left(a+b+c\right)^2=3-\dfrac{1}{3}\cdot\dfrac{9}{4}=\dfrac{9}{4}=VP\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{2}\)
\(VT=\left(\dfrac{a}{1+4c^2}+\dfrac{b}{1+4a^2}+\dfrac{c}{1+4b^2}\right)+\left(\dfrac{1}{1+4c^2}+\dfrac{1}{1+4a^2}+\dfrac{1}{1+4b^2}\right)\)
\(VT=\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)+3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)
Xét \(\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2a}{1+4c^2}\le\dfrac{4c^2a}{4c}=ca\\\dfrac{4a^2b}{1+4a^2}\le\dfrac{4a^2b}{4a}=ab\\\dfrac{4b^2c}{1+4b^2}\le\dfrac{4b^2c}{4b}=bc\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\ge\dfrac{3}{2}-\left(ab+bc+ca\right)\) (1)
Xét \(3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2}{1+4c^2}\le\dfrac{4c^2}{4c}=c\\\dfrac{4a^2}{1+4a^2}\le\dfrac{4a^2}{4a}=a\\\dfrac{4b^2}{1+4b^2}\le\dfrac{4b^2}{4b}=b\end{matrix}\right.\)
\(\Rightarrow3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\ge\dfrac{3}{2}\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge\dfrac{3}{2}-\left(ab+bc+ca\right)+\dfrac{3}{2}\)
\(\Rightarrow VT\ge3-\left(ab+bc+ca\right)\) (3)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{3}{4}\ge ab+bc+ca\)
\(\Rightarrow3-\dfrac{3}{4}\le3-\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{9}{4}\le3-\left(ab+bc+ca\right)\) (4)
Từ (3) và (4)
\(\Rightarrow VT\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+4c^2}\ge\dfrac{9}{4}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a\left(b+c\right)}{4}\ge2\sqrt{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a\left(b+c\right)}{4}=2\sqrt{\dfrac{1}{4a^2}=\dfrac{1}{a}=\dfrac{abc}{a}=bc}}\)
Tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b\left(c+a\right)}{4}\ge\dfrac{1}{b}=ac\)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c\left(a+b\right)}{4}\ge\dfrac{1}{c}=ab\)
Cộng theo vế:
\(\Rightarrow VT+\dfrac{ab+bc+ac}{2}\ge ab+bc+ac\)
\(\Rightarrow VT\ge\dfrac{ab+bc+ac}{2}\)
Tiếp tục áp dụng AM-GM: \(ab+bc+ac\ge3^3\sqrt{a^2b^2c^2}=3\)
\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Dấu bằng xảy ra khi a=b=c=1