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a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
\(\le2\cdot\left(1+1+1\right)\left(a+b+c\right)\le6\)
\(\Rightarrow VT^2\le6\Rightarrow VT\le\sqrt{6}=VP\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
b)Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{a+\sqrt{b+\sqrt{2c}}}+\sqrt{b+\sqrt{c+\sqrt{2a}}}+\sqrt{c+\sqrt{a+\sqrt{2b}}}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+c+Σ\sqrt{b+\sqrt{2c}}\right)\)
\(=3\left(6+\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)\)
Đặt \(A^2=\left(\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)^2\)
\(\le\left(1+1+1\right)\left(a+b+c+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\)
\(=3\left(6+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\)
Đặt tiếp: \(B^2=\left(\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)^2\)
\(\le2\cdot\left(1+1+1\right)\left(a+b+c\right)\le36\Rightarrow B\le6\)
\(\Rightarrow A^2\le3\left(6+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\le3\cdot12=36\Rightarrow A\le6\)
\(\Rightarrow VT^2\le3\left(6+\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)\)
\(\le3\left(6+6\right)=3\cdot12=36\Rightarrow VT\le6=VP\)
Xảy ra khi \(a=b=c=2\)
Áp dụng BĐT Bunhiacopxki, ta có :
\(\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\)
\(\Rightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le6\left(a+b+c\right)\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)
Easy!
\(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
\(=\sqrt{\frac{3}{2}}\left[\sqrt{\left(a+b\right).\frac{2}{3}}+\sqrt{\left(b+c\right).\frac{2}{3}}+\sqrt{\left(c+a\right).\frac{2}{3}}\right]\) (*)
Áp dụng BĐT Cô si ngược,ta có:
(*) \(\le\sqrt{\frac{3}{2}}\left[\frac{a+b+\frac{2}{3}}{2}+\frac{b+c+\frac{2}{3}}{2}+\frac{c+a+\frac{2}{3}}{2}\right]\)
\(=\sqrt{\frac{3}{2}}\left(a+b+c+1\right)=\sqrt{\frac{3}{2}}.2=\sqrt{6}^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a+b=b+c=c+a=\frac{2}{3}\\a+b+c=1\end{cases}\Leftrightarrow}a=b=c=\frac{1}{3}\)
Đề đánh bị lỗi.
Áp dụng bất đẳng thức Bunhiacopski:
\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{\left[\sqrt{c}^2+\sqrt{\left(a-c\right)}^2\right]\left[\sqrt{c}^2+\sqrt{\left(b-c\right)}^2\right]}\)
\(=\sqrt{\left(c+a-c\right)\left(c+b-c\right)}=\sqrt{ab}\)
sửa đề\(\frac{1}{x^2+1}+\frac{1}{y^2+1}\ge\frac{2}{1+xy}\)
\(\Leftrightarrow\frac{1}{x^2+1}+\frac{1}{y^2+1}-\frac{2}{1+xy}\ge0\)
\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\frac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\frac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)( luôn đúng với \(x,y\ge1\))
Đpcm
bđt cần c/m tương đương với:
\(\left(\frac{b+c}{\sqrt{a}}+\sqrt{a}\right)+\left(\frac{a+c}{\sqrt{b}}+\sqrt{b}\right)+\left(\frac{a+b}{\sqrt{c}}+\sqrt{c}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+3\\ \ \)\(\left(a+b+c\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+3\)
Mặt khác:
\(a+b+c\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3}\)
\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\frac{9}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
=> \(VT\ge3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
Ta cần c/m:
\(3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+3\)
<=> \(\sqrt{a}+\sqrt{b}+\sqrt{c}\ge3\sqrt[3]{\sqrt{abc}}=3\)(BĐt Cô-si)
xong rồi bạn nhé
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
Áp dụng BĐT Bunhiakovski
\(VT^2=\left(\sqrt{a+b}.1+\sqrt{b+c}.1+\sqrt{c+a}.1\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)\)
\(=3.2\left(a+b+c\right)=6\)
Do đó \(VT\le\sqrt{6}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{a+b}}{1}=\dfrac{\sqrt{b+c}}{1}=\dfrac{\sqrt{c+a}}{1}\\a+b+c=1\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)