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Áp dụng BĐT Cauchy cho các số không âm, ta có:
\(\dfrac{c}{a}+\dfrac{b}{c}\ge2\sqrt{\dfrac{c}{a}\cdot\dfrac{b}{c}}=2\sqrt{\dfrac{b}{a}}\)
\(\dfrac{b}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{b}{a}\cdot\dfrac{a}{b}}=2\)
Vì \(2\sqrt{\dfrac{b}{a}}\ge2\) nên \(\dfrac{c}{a}+\dfrac{b}{c}\ge\dfrac{b}{a}+\dfrac{a}{b}\) (đpcm)
Áp dụng BĐT Cauchy cho các số không âm , ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\) ≥ \(2\sqrt{\dfrac{a}{b}.\dfrac{b}{c}}=2\sqrt{\dfrac{a}{c}}\left(1\right)\)
\(\dfrac{b}{c}+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{b}{c}.\dfrac{c}{a}}=2\sqrt{\dfrac{b}{a}}\left(2\right)\)
\(\dfrac{a}{b}+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{a}{b}.\dfrac{c}{a}}=2\sqrt{\dfrac{c}{b}}\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3) , ta có :
\(2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) ≥ \(2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)
⇔ \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) ≥ \(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Lời giải:
Từ \(a+b+c\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow a+b+c\geq \frac{ab+bc+ac}{abc}\Rightarrow abc(a+b+c)\geq ab+bc+ac\)
\(\Rightarrow a^2b^2c^2(a+b+c)^2\geq (ab+bc+ac)^2(1)\)
Áp dụng BĐT AM-GM:
\(a^2b^2+b^2c^2\geq 2ab^2c\)
\(b^2c^2+c^2a^2\geq 2abc^2\)
\(a^2b^2+c^2a^2\geq 2a^2bc\)
Cộng theo vế, rút gọn \(\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)\)
\(\Rightarrow (ab+bc+ac)^2\geq 3abc(a+b+c)(2)\)
Từ \((1);(2)\Rightarrow a^2b^2c^2(a+b+c)^2\geq 3abc(a+b+c)\)
\(\Rightarrow abc(a+b+c)\geq 3\Rightarrow a+b+c\geq \frac{3}{abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)
ok thỏa thuận rồi tui làm nửa sau thui nhé :D
Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:
\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)
Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:
\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)
\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)
Can you continue
Đặt \(A=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}\)
\(=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{c}{b}\)
Áp dụng bất đẳng thức :
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\dfrac{b}{a}+\dfrac{a}{b}\ge2\)
\(\Rightarrow\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{b}\ge6\)
Đặt \(B=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\)
\(\Rightarrow B+3=\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1\)
\(=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
Ta có : \(2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
\(\Rightarrow B+3\ge\dfrac{9}{2}\Rightarrow B\ge\dfrac{3}{2}\)
\(\Rightarrow A+B\ge\dfrac{15}{2}\)
Dấu " = " xảy ra khi a = b = c .
\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)
Lời giải:
Áp dụng BĐT Svac-xơ ta có:
\(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}=\frac{1^2}{a}+\frac{2^2}{b}+\frac{3^2}{c}\geq \frac{(1+2+3)^2}{a+b+c}=\frac{36}{a+b+c}\)
Ta có đpcm
Dấu "=" xảy ra khi \(\frac{1}{a}=\frac{2}{b}=\frac{3}{c}\)