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Áp dụng BĐT Cô - Si , ta có :
\(\dfrac{a}{b^2}+\dfrac{1}{a}\) ≥ \(2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=2.\dfrac{1}{b}\left(a,b>0\right)\left(1\right)\)
\(\dfrac{b}{c^2}+\dfrac{1}{b}\text{ ≥ }2\sqrt{\dfrac{b}{c^2}.\dfrac{1}{b}}=2.\dfrac{1}{c}\left(b,c>0\right)\left(2\right)\)
\(\dfrac{c}{a^2}+\dfrac{1}{c}\text{≥}2\sqrt{\dfrac{c}{a^2}.\dfrac{1}{c}}=2.\dfrac{1}{a}\left(a,c>0\right)\left(3\right)\)
Từ ( 1 ; 2 ; 3) Ta có :
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
⇔\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\) ≥ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Bài 3:
a) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\) \(\geq 2.\frac{(1+1)^2}{2xy+x^2+y^2}=\frac{8}{(x+y)^2}=8\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{1}{x^2+y^2}=\frac{1}{2xy}+\left (\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)\geq \frac{1}{2xy}+\frac{(1+1)^2}{2xy+x^2+y^2}\)
\(=\frac{1}{2xy}+\frac{4}{(x+y)^2}\)
Theo BĐT AM-GM:
\(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{1}{2xy}\geq 2\)
Do đó \(\frac{1}{xy}+\frac{1}{x^2+y^2}\geq 2+4=6\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
Bài 1: Thiếu đề.
Bài 2: Sai đề, thử với \(x=\frac{1}{6}\)
Bài 4 a) Sai đề với \(x<0\)
b) Áp dụng BĐT AM-GM:
\(x^4-x+\frac{1}{2}=\left (x^4+\frac{1}{4}\right)-x+\frac{1}{4}\geq x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x^4=\frac{1}{4}\\ x=\frac{1}{2}\end{matrix}\right.\) (vô lý)
Do đó dấu bằng không xảy ra , nên \(x^4-x+\frac{1}{2}>0\)
Bài 6: Áp dụng BĐT AM-GM cho $6$ số:
\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^3b^3c^3d^3}=6\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=d=1\)
5) a) Đặt b+c-a=x;a+c-b=y;a+b-c=z thì 2a=y+z;2b=x+z;2c=x+y
Ta có:
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
Vậy ta suy ra đpcm
b) Ta có: a+b>c;b+c>a;a+c>b
Xét: \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
.Tương tự:
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c};\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy ta có đpcm
6) Ta có:
\(a^2+b^2+c^2+d^2+ab+cd\ge2ab+2cd+ab+cd=3\left(ab+cd\right)\)
\(ab+cd=ab+\dfrac{1}{ab}\ge2\)
Suy ra đpcm
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{b}=\dfrac{2}{c}-\dfrac{1}{a}=\dfrac{2a-c}{ac}\\\dfrac{1}{a}=\dfrac{2}{c}-\dfrac{1}{b}=\dfrac{2b-c}{bc}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a-c=\dfrac{ac}{b}\\2b-c=\dfrac{bc}{a}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+c}{2a-c}=\dfrac{b\left(a+c\right)}{ac}=\dfrac{ab}{ac}+\dfrac{bc}{ac}=\dfrac{b}{c}+\dfrac{b}{a}\\\dfrac{b+c}{2b-c}=\dfrac{a\left(b+c\right)}{bc}=\dfrac{ab}{bc}+\dfrac{ac}{bc}=\dfrac{a}{c}+\dfrac{a}{b}\end{matrix}\right.\)
Áp dụng bđt Cosi cho 2 số sương ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}=0\Leftrightarrow\dfrac{2}{c}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{c}\ge2\)(áp dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\))
Ta có: \(\dfrac{a+c}{2a-c}+\dfrac{b+c}{2b-c}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{a+b}{c}\ge2+2=4\)
Dấu "=" xawy ra khi và chỉ khi a=b=c
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
Đặt P=\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Không mất tính tổng quát giả sử a ≥b ≥ c , thế thì \(\dfrac{1}{b+c}\ge\dfrac{1}{c+a}\ge\dfrac{1}{a+b}\) .Áp dụng bất đẳng thức Chebyshev cho hai dãy đơn điệu cùng chiều ta có :
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{1}{3}\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\left(\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\right)\)
Hay \(P\ge\dfrac{1}{3}\left(P+3\right)\) nghĩa là \(P\ge\dfrac{3}{2}^{\left(đpcm\right)}\)
Áp dụng BĐT Cô si dạng phân số ta có :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
=> ĐPCM .
b) Vì a,b,c > 0 .
Áp dụng BĐT Cô si ta có :
\(\dfrac{a^2}{b}+b\ge2a\) (1)
Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2b\) (2)
\(\dfrac{c^2}{a}+a\ge2c\) (3)
Cộng từng vế => ĐPCM .
Ta có:\(\dfrac{ab}{a+b}=\dfrac{ab+b^2-b^2}{a+b}=\dfrac{b\left(a+b\right)-b^2}{a+b}=b-\dfrac{b^2}{a+b}\)
Tương tự với các vế ta được:
\(\dfrac{bc}{b+c}=c-\dfrac{c^2}{b+c}\) và \(\dfrac{ac}{a+c}=a-\dfrac{a^2}{a+c}\)
Cộng theo vế:
\(VT=a+b+c-\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\right)\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{1}{2}\left(a+b+c\right)\)
Bài này áp dụng BĐT Cauchy (Cô-si) cho 2 số.
Ta có: a^2/b + b >= 2.căn[(a^2/b).b] = 2.căn(a^2) = 2|a| >= 2a
Tương tự, b^2/c + c >= 2|b| >= 2b
................c^2/a + a >= 2|c| >= 2c
Cộng vế với vế, ta được:
a^2/b + b^2/c + c^2/a + a + b + c >= 2a + 2b + 2c
<=> a^2/b + b^2/c + c^2/a >= a + b + c (điều phải chứng minh)
Xét : \(\dfrac{a^2}{b}+b=\dfrac{a^2+b^2}{b}\ge\dfrac{2ab}{b}=2a\)
\(\Rightarrow\dfrac{a^2}{b}+b\ge2a\)
Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2a;\dfrac{c^2}{a}+a\ge2c\)
Cộng theo vế cac BPT trên :
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)