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\(a^5+b^2+ab+6\ge3a^2b+6\)
\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)
\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)
\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)
\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
\(\Rightarrow P\le\sqrt{1}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: \(\dfrac{1}{4-\sqrt{ab}}\le\dfrac{1}{4-\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(\left(a^2+b^2;b^2+c^2;c^2+a^2\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x+y+z=6\\x;y;z>0\end{matrix}\right.\)
Làm nốt :v
Lời giải:
Từ \(ab+bc+ac=1\Rightarrow a^2+ab+bc+ac=a^2+1\)
\(\Leftrightarrow (a+b)(a+c)=a^2+1\)
Tương tự: \(\left\{\begin{matrix} b^2+1=(b+c)(b+a)\\ c^2+1=(c+a)(c+b)\end{matrix}\right.\)
Khi đó:
\(A=\frac{(b^2+bc)(c^2+ca)(a^2+ab)}{\sqrt{(a^4+a^2)(b^4+b^2)(c^4+c^2)}}\) \(=\frac{b(b+c)c(c+a)a(a+b)}{\sqrt{a^2b^2c^2(a^2+1)(b^2+1)(c^2+1)}}\)
\(=\frac{abc(a+b)(b+c)(c+a)}{abc\sqrt{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}}\) \(=\frac{abc(a+b)(b+c)(c+a)}{abc(a+b)(b+c)(c+a)}=1\)
Vậy \(A=1\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\end{matrix}\right.\)\(\Rightarrow x+y+z=xyz\)
\(\Rightarrow P=xy+yz+xz-\sqrt{x^2+1}-\sqrt{y^2+1}-\sqrt{z^2+1}\)
Khi \(a=b=c=\frac{1}{\sqrt{3}}\Rightarrow x=y=z=\sqrt{3}\Rightarrow P=3\)
Ta sẽ chứng minh \(P=3\) là giá tri nhỏ nhất của \(P\)
\(\Rightarrow BDT\Leftrightarrow xy+yz+xz-3\ge\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\)
Ta có BĐT \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2\)
\(\Leftrightarrow\left(xy+yz+xz\right)^2\ge x^2y^2z^2+2xyz\left(x+y+z\right)\)\(=3\left(x+y+z\right)^2\)
Xét \(VT^2=\left(xy+yz+xz-3\right)^2=\left(xy+yz+xz\right)^2-6\left(xy+yz+xz\right)+9\)
\(\ge3\left(x+y+z\right)^2-6\left(xy+yz+xz\right)+9\)\(=3\left(x^2+y^2+z^2\right)+9\left(1\right)\)
Và \(VP^2\le\left(1+1+1\right)\left(x^2+y^2+z^2+3\right)=3\left(x^2+y^2+z^2\right)+9\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\) ta có ĐPCM. Vậy \(P_{min}=3\Rightarrow a=b=c=\frac{1}{\sqrt{3}}\)
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
Ta có:
1+a2 = ab+bc+ca+a2 = a(a+b)+c(a+b)=(a+b)(a+c)
Tương tự: 1+b2 = (b+c)(b+a)
1+c2 = (c+a)(c+b)
\(\Rightarrow\) P = \(2a\sqrt{\dfrac{1}{\left(a+b\right)\left(a+c\right)}}+2b\sqrt{\dfrac{1}{\left(b+c\right)\left(b+a\right)}}+2c\sqrt{\dfrac{1}{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT Cô-si ta có:
P\(\le\)\(a\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+b\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{b+a}\right)+c\left(\dfrac{1}{4\left(c+b\right)}+\dfrac{1}{c+a}\right)\)\(\le\)\(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4\left(b+c\right)}+\dfrac{b}{b+a}+\dfrac{c}{4\left(c+b\right)}+\dfrac{c}{c+a}\)
= \(\dfrac{1}{4}+2=\dfrac{9}{4}\)
\(\Rightarrow\)Pmin = \(\dfrac{9}{4}\)
Dấu "=" xảy ra\(\Leftrightarrow\) b=c=\(\dfrac{a}{7}\)=\(\dfrac{\sqrt{15}}{15}\) \(\Rightarrow\) a = \(\dfrac{7\sqrt{15}}{15}\)
Lời giải:
Vì $abc=1$ nên tồn tại $x,y,z$ sao cho : \((a,b,c)=\left(\frac{x}{y}, \frac{y}{z}, \frac{z}{x}\right)\)
Khi đó:
\(\text{VT}=\frac{1}{\sqrt{\frac{x}{z}+\frac{x}{y}+2}}+\frac{1}{\sqrt{\frac{y}{x}+\frac{y}{z}+2}}+\frac{1}{\sqrt{\frac{z}{y}+\frac{z}{x}+2}}=\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}+\frac{\sqrt{xz}}{\sqrt{xy+yz+2xz}}+\frac{\sqrt{xy}}{\sqrt{xz+yz+2xy}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}^2\leq (1+1+1)\left(\frac{yz}{xy+xz+2yz}+\frac{xz}{xy+yz+2xz}+\frac{xy}{xz+yz+2xy}\right)\)
\(\leq 3\left[\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)+\frac{xz}{4}\left(\frac{1}{xy+xz}+\frac{1}{xz+yz}\right)+\frac{xy}{4}\left(\frac{1}{xz+xy}+\frac{1}{yz+xy}\right)\right]\)
hay \(\text{VT}^2\leq \frac{3}{4}.\left(\frac{xy+yz}{xy+yz}+\frac{xy+xz}{xy+xz}+\frac{yz+xz}{yz+xz}\right)=\frac{9}{4}\)
\(\Rightarrow \text{VT}\leq \frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$
Đề bài sai, bạn kiểm tra lại điều kiện \(a^2+b^2+c^2=1\)