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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)
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
\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(S=\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{a}{b}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{ac}{ca}}=2\\\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{bc}{cb}}=2\\\frac{b}{a}+\frac{a}{b}\ge2\sqrt{\frac{ab}{ba}}=2\end{cases}}\)
\(\Rightarrow\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{a}{b}\right)\ge2+2+2=6\)
\(\Leftrightarrow\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge6\)
\(\Leftrightarrow S\ge6\left(đpcm\right)\)
\(\Rightarrow S_{min}=6\)
Dấu " = " xảy ra khi \(a=b=c\)
Chúc bạn học tốt !!!
\(a^2+b^2+c^2\ge ab+bc+ca=2\)
Áp dụng BĐT C-S:
\(P\ge\dfrac{\left(a+b+c\right)^2}{3-\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2+4}{3-\left(a^2+b^2+c^2\right)}\)
Đặt \(a^2+b^2+c^2=x\)
Ta cần c/m: \(\dfrac{x+4}{3-x}\ge6\Leftrightarrow x+4\ge18-6x\)
\(\Leftrightarrow x\ge2\) (đúng)
Dấu = xảy ra khi \(a=b=c=\pm\sqrt{\dfrac{2}{3}}\)
\(a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
<=> \(a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{ab+bc+ca}{1}=ab+bc+ca\) (thay abc = 1)
=> a + b + c - ab - bc - ca = 0
<=> 1 + a + b + c - ab - bc - ca - 1 = 0
<=> abc + a + b + c - ab - bc - ca - 1 = 0 (thay 1 = abc)
<=> (abc - ab) + (b - bc) + (a - ca) + (c - 1) = 0
<=> ab(c - 1) - b(c - 1) - a(c - 1) + (c - 1) = 0
<=> (c - 1)(ab - b - a + 1) = 0
<=> (c - 1)[b(a - 1) - (a - 1)] = 0
<=> (c - 1)(a - 1)(b - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\) (đpcm)
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Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
\(S=\dfrac{a}{a+b+c}+\dfrac{a+b+c}{a}+\dfrac{b}{a+b+c}+\dfrac{a+b+c}{b}+\dfrac{c}{a+b+c}+\dfrac{a+b+c}{c}-\dfrac{a}{b}-\dfrac{a}{c}-\dfrac{b}{a}-\dfrac{b}{c}-\dfrac{c}{a}-\dfrac{c}{b}=\dfrac{a+b+c}{a+b+c}+\dfrac{a+b+c-b-c}{a}+\dfrac{a+b+c-a-c}{b}+\dfrac{a+b+c-a-b}{c}=1+1+1+1=4\)
\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(S=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\)
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 \(\left\{{}\begin{matrix}\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\end{matrix}\right.\)\(\Rightarrow a=b=c\)