\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)

=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)

=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)

Dấu = xảy ra khi x=y=z=6căn 2

\(a^2+b^2=\left(a+b-c\right)^2=a^2+\left(b-c\right)^2+2a\left(b-c\right)=b^2+\left(a-c\right)^2+2b\left(a-c\right)\)

\(\Rightarrow\left\{{}\begin{matrix}b^2=\left(b-c\right)^2+2a\left(b-c\right)\\a^2=\left(a-c\right)^2+2b\left(a-c\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\dfrac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}\)

\(=\dfrac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\dfrac{a-c}{b-c}\) (đpcm)

Lời giải:

Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:

\(ab+bc+ac=1\) và cần phải CMR:

\(P=\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}+\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}+\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}\)

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Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)

\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)

\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)

\(=(b+c)\sqrt{(a+b)(a+c)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}(1)\)

Tương tự:

\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)

\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)

Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)

\(=0\)

Ta có đpcm.

Ta có: \(1 + x^3 = (1 + x)(x^2 - x + 1)\)

=>\(\sqrt{1 + x^3} = \sqrt{(1 + x)(x^2 - x + 1)} \le \frac{(1 + x) + (x^2 - x + 1)}{2} = \frac{x^2 + 2}{2}\) (1)

Chứng minh tương tự, ta cũng có:

\(\sqrt{1 + y^3}\le\frac{y^2 + 2}{2}\left(2\right);\sqrt{1+z^3}\le\frac{z^2+2}{2}\left(3\right)\)

Từ (1),(2) suy ra \(\sqrt{(1 + x^3)(1 + y^3)} \le \frac{(x^2 + 2)(y^2 + 2)}{4}\)

=>\(\frac{x^2}{\sqrt{(1+x^3)(1+y^3)}} \ge \frac{4x^2}{(x^2+2)(y^2+2)}\)

Chứng minh tương tự, ta cũng có:

\(\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}\ge\frac{4y^2}{\left(y^2+2\right)\left(z^2+2\right)}\)

\(\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}>=\frac{4z^2}{\left(z^2+2\right)\left(x^2+2\right)}\)

=>\(P \ge \frac{4x^2}{(x^2+2)(y^2+2)} + \frac{4y^2}{(y^2+2)(z^2+2)} + \frac{4z^2}{(z^2+2)(x^2+2)}\)

=>\(P \ge \frac{4 \left[ x^2(z^2+2) + y^2(x^2+2) + z^2(y^2+2) \right]}{(x^2+2)(y^2+2)(z^2+2)}\)

=>\(P \ge \frac{4(x^2y^2 + y^2z^2 + z^2x^2 + 2x^2 + 2y^2 + 2z^2)}{(x^2+2)(y^2+2)(z^2+2)}\)

Đặt \(a = x^2, b = y^2, c = z^2\)

=>\(abc=\left(xyz\right)^2=64\)

\((a+2)(b+2)(c+2) = abc + 2(ab+bc+ca) + 4(a+b+c) + 8\)

\(=64+8+2(ab+bc+ca)+4(a+b+c)\)

\(=72+2(ab+bc+ca)+4(a+b+c)\)

\(4(ab + bc + ca + 2a + 2b + 2c) = 2 \cdot \left[ 2(ab+bc+ca) + 4(a+b+c) \right]\)

Đặt \(S = 2(ab+bc+ca) + 4(a+b+c)\)

=>\(P \ge \frac{2S}{72 + S}\)

Lấy \(S_1, S_2 \in (0; +\infty)\) sao cho 0<S1<S2

\(f(S_2) - f(S_1) = \left( 2 - \frac{144}{72+S_2} \right) - \left( 2 - \frac{144}{72+S_1} \right)\)

\(= \frac{144}{72+S_1} - \frac{144}{72+S_2}\)

\(=144\cdot\frac{(72+S_2) - (72+S_1)}{(72+S_1)(72+S_2)}=\frac{144\cdot\left(S_2-S_1\right)}{\left(72+S_1\right)\left(72+S_2\right)}\) >0

=>f(S2)>f(S1)

=>f(S) đồng biến trên khoảng (0;+∞)

S>=144

=>\(f(S) \ge f(144) = \frac{2 \times 144}{72 + 144} = \frac{288}{216} = \frac{4}{3}\)

=>P>=4/3

Dấu '=' xảy ra khi x=y=z=2

Đề bài sai, biểu thức này ko có min

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\dfrac{1}{\sqrt{x}+2\sqrt{y}}\le\dfrac{1}{9}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{y}}\right)\)

Tương tự cho 2 BĐT trên ta có:

\(\dfrac{1}{3}VP\le\dfrac{1}{9}\cdot3\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)=\dfrac{1}{3}VT\)

Xảy ra khi \(x=y=z\)