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1.VT= \(\dfrac{x}{z}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{x}{y}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)\)
Áp dụng BĐT Cô-si cho 2 số dương, ta có:
\(\dfrac{x}{y}+\dfrac{y}{x}\)≥ 2\(\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}\)=2; tương tự \(\dfrac{x}{z}+\dfrac{z}{x}\)≥2; \(\dfrac{y}{z}+\dfrac{z}{y}\)≥2.
Cộng 3 BĐT trên, ta được đpcm.
2.Đặt b+c-a= x, a+c-b= y, a+b-c= z. Khi đó x,y,z>0.
2a= y+z; 2b= x+z; 2c= x+y. Khi đó bđt cần chứng minh trở thành:
\(\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}\)≥6.
Theo bài 1 bđt luôn đúng
\(\left\{{}\begin{matrix}\dfrac{a}{b+c}>\dfrac{a}{a+b+c}\\\dfrac{b}{c+a}>\dfrac{b}{a+b+c}\\\dfrac{c}{a+b}>\dfrac{c}{a+b+c}\end{matrix}\right.\Rightarrow\dfrac{a}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{a+b}>\dfrac{a+b+c}{a+b+c}=1\)
\(\left\{{}\begin{matrix}\dfrac{a}{b+c}< \dfrac{2a}{a+b+c}\\\dfrac{b}{c+a}< \dfrac{2b}{a+b+c}\\\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\end{matrix}\right.\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< \dfrac{2a+2b+2c}{a+b+c}=2\)
Từ trên \(\Rightarrowđpcm\)
Bài 1:a,b,c ba cạnh tam giác => a,b,c dương
\(\left\{{}\begin{matrix}a+c>b\\a+b>c\\b+c>a\end{matrix}\right.\) ta có: \(\dfrac{x}{y}< \dfrac{x+p}{y+p}\forall_{x,y,p>0\&x< y}\)
\(VT=\dfrac{a}{a+b}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+c}{a+b}+\dfrac{b}{c+a}< \dfrac{a+c+c}{a+b+c}+\dfrac{b+b}{a+b+c}=\)
\(=\dfrac{a+b+c+b+c}{a+b+c}< \dfrac{\left(a+b+c\right)+\left(A+b+c\right)}{a+b+c}< \dfrac{2\left(b+a+c\right)}{a+b+c}=2=VP\)
p/s: đề sao làm vậy:
mình nghi đề phải thế này: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) cách làm đơn giản hơn
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
Áp dụng BĐT AM-GM ( one shot là done )
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}}=3\sqrt[3]{1}=3\)
Đẳng thức xảy ra khi \(a=b=c\)
Đề sai nhé . \(\ge3\)
Đặt \(b+c-a=x;a+c-b=y;a+b-c=z\) ( x ; y ; z luôn > 0 )
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2};b=\dfrac{x+z}{2}\end{matrix}\right.\)
Ta có : \(A=\dfrac{y+z}{2x}+\dfrac{x+z}{2y}+\dfrac{x+y}{2z}\)
Áp dụng BĐT Cô - si cho 3 số , ta có :
\(A\ge3\sqrt[3]{\dfrac{\left(y+z\right)\left(x+z\right)\left(x+y\right)}{2x.2y.2z}}\ge3\sqrt[3]{\dfrac{8xyz}{8xyz}}=3\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)
\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(S=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{b}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{ab}{ba}}=2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2+2+2=6\)
\(\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
\(\Leftrightarrow S\ge6\) ( đpcm )
\(\Rightarrow S_{min}=6\)
Dấu " = " xảy ra khi \(a=b=c\)
cách 1 sử dụng BĐT
a)
\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\right)\)đã áp cô_si --> áp tới bến luôn
\(S=\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\right)\ge6\sqrt[6]{\dfrac{\left(abc\right)^2}{\left(abc\right)^2}}=6\) =>dpcm
b) min S=6
khi \(\dfrac{a}{b}=\dfrac{b}{a}=\dfrac{c}{a}=\dfrac{a}{c}=\dfrac{b}{c}=\dfrac{c}{b}\Rightarrow a=b=c\)
cách2sử dụng HĐT \(\left(x-y\right)^2\ge0\forall x,y\)
\(S=\left(\dfrac{a}{b}-2+\dfrac{b}{a}\right)+\left(\dfrac{c}{b}-2+\dfrac{b}{c}\right)+\left(\dfrac{a}{c}-2+\dfrac{c}{a}\right)+6\)
\(S=\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)^2+\left(\sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a}}\right)^2+\left(\sqrt{\dfrac{a}{c}}-\sqrt{\dfrac{c}{a}}\right)^2+6\ge6\)=> dpcm
Min S=6
khi \(\left\{{}\begin{matrix}\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\end{matrix}\right.\)\(\Rightarrow a=b=c\)
Ta có bất đẳng thức sau
a2 + b2 + c2 \(\ge\) ab + bc + ca (1)
Dấu "=" xảy ra <=> a = b = c
Thật vậy (1) <=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca \(\ge0\)
<=> (a - b)2 + (b - c)2 + (c - a)2 \(\ge0\) (bđt này luôn đúng)
Khi đó ta được (1) <=> 2(a2 + b2 + c2) \(\ge\) 2(ab + bc + ca)
<=> 3(a2 + b2 + c2) \(\ge\) 2ab + 2bc + 2ca + a2 + b2 + c2
<=> 3(a2 + b2 + c2) \(\ge\) (a + b + c)2
=> -(a2 + b2 + c2) \(\le\dfrac{(a+b+c)^2}{3}\)
Ta có \(P=\dfrac{b+c}{b+c-a}+\dfrac{c+a}{c+a-b}+\dfrac{a+b}{a+b-c}\)
\(=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}+3\)
\(=\dfrac{a^2}{ab+ac-a^2}+\dfrac{b^2}{ab+bc-b^2}+\dfrac{c^2}{ac+bc-c^2}+3\)
\(\ge\dfrac{\left(a+b+c\right)^2}{ab+ac-a^2+ab+bc-b^2+ac+bc-c^2}+3\) (BĐT Schwarz)
\(=\dfrac{\left(a+b+c\right)^2}{2ab+2ac+2bc-a^2-b^2-c^2}+3\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2-2\left(a^2+b^2+c^2\right)}+3\)
\(\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2-\dfrac{2}{3}\left(a+b+c\right)^2}+3=\dfrac{1}{1-\dfrac{2}{3}}+3=6\) (đpcm)