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\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(S=\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{a}{b}\right)\)
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 !!!
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
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:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
\(a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
<=> \(a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{ab+bc+ca}{1}=ab+bc+ca\) (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 : Do a ; b ; c là 3 cạnh của 1 tam giác nên :
\(\dfrac{a}{a+b+c}< \dfrac{a}{b+c}< \dfrac{2a}{a+b+c}\)
\(\dfrac{b}{a+b+c}< \dfrac{b}{c+a}< \dfrac{2b}{a+b+c}\)
\(\dfrac{c}{a+b+c}< \dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\)
Cộng 3 vế với nhau , ta có :
\(1< \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\left(đpcm\right)\)
Ta có :
\(\dfrac{â}{b+c}>\dfrac{a}{a+b+c}\);
\(\dfrac{b}{c+a}>\dfrac{b}{a+b+c}\);
\(\dfrac{c}{a+b}>\dfrac{c}{a+b+c}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}>\dfrac{a+b+c}{a+b+c}=1\) (*)
Ta có bất đằng thức tam giác : a+b > c ; b+c > a ; a+c > b
\(\Rightarrow\dfrac{a}{b+c}< 1;\dfrac{b}{a+c}< 1;\dfrac{c}{a+b}< 1\)
Vì \(\dfrac{a}{b+c}< 1\Rightarrow\dfrac{a}{b+c}< \dfrac{a+a}{a+b+c}=\dfrac{2a}{a+b+c}\)
Tương tự :
\(\dfrac{b}{a+c}< \dfrac{2b}{a+b+c};\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\) (**)
Kết hợp (*) với (**)
=> ĐPCM
ta có:\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
=>\(\left[\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right].\left(a+b+c\right)=a+b+c\)
=>\(\dfrac{a^2}{c+b}+\dfrac{ab}{a+c}+\dfrac{ac}{a+b}+\dfrac{b^2}{a+c}+\dfrac{ba}{c+d}+\dfrac{bc}{a+b}+\dfrac{ca}{c+d}+\dfrac{cb}{a+c}+\dfrac{c^2}{a+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{c+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
chúc bạn học tốt ^ ^
\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(S=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\)
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 \(\left\{{}\begin{matrix}\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\\\left(\sqrt{\dfrac{c}{b}}-\sqrt{\dfrac{b}{c}}\right)=0\end{matrix}\right.\)\(\Rightarrow a=b=c\)