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Lời giải:
Ta có:
\(A=\sqrt{(x+y)(y+z)(z+x)}\left(\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\right)\)
\(A=\frac{(y+z)\sqrt{(x+y)(x+z)}}{x}+\frac{(z+x)\sqrt{(y+z)(y+x)}}{y}+\frac{(x+y)\sqrt{(z+x)(z+y)}}{z}\)
Áp dụng BĐT Bunhiacopxky:
\((x+y)(x+z)\geq (x+\sqrt{yz})^2\) và tương tự với những biểu thức khác suy ra:
\(A\geq \frac{(y+z)(x+\sqrt{yz})}{x}+\frac{(z+x)(y+\sqrt{xz})}{y}+\frac{(x+y)(z+\sqrt{xy})}{z}\)
hay \(A\geq 2(x+y+z)+\frac{(y+z)\sqrt{yz}}{x}+\frac{(z+x)\sqrt{zx}}{y}+\frac{(x+y)\sqrt{xy}}{z}\)
hay \(A\geq 2(x+y+z)+\underbrace{\frac{yz(y+z)\sqrt{yz}+xz(x+z)\sqrt{xz}+xy(x+y)\sqrt{xy}}{xyz}}_{M}\)
Đặt \((x,y,z)=(a^2,b^2,c^2)\)
Khi đó: \(M=\frac{a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(a^2+c^2)}{a^2b^2c^2}\)
Áp dụng BĐT AM-GM:
\(a^5b^3+a^3b^5\geq 2\sqrt{a^8b^8}=2a^4b^4\)
\(b^5c^3+c^5b^3\geq 2b^4c^4\)
\(c^5a^3+a^5c^3\geq 2c^4a^4\)
\(\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2(a^4b^4+b^4c^4+c^4a^4)\) (1)
(cộng các BĐT theo vế)
Tiếp tục AM-GM:
\(a^4b^4+b^4c^4\geq 2a^2b^4c^2; b^4c^4+c^4a^4\geq 2a^2b^2c^4; c^4a^4+a^4b^4\geq 2a^4b^2c^2\)
\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^2c^2(a^2+b^2+c^2)\) (2)
Từ\((1); (2)\Rightarrow a^3b^3(a^2+b^2)+b^3c^3(b^2+c^2)+c^3a^3(c^2+a^2)\geq 2a^2b^2c^2(a^2+b^2+c^2)\)
\(\Rightarrow M\geq 2(a^2+b^2+c^2)=2(x+y+z)\)
Do đó: \(A\geq 2(x+y+z)+M\geq 4(x+y+z)\Leftrightarrow A\geq 4\sqrt{2}\)
Vậy \(A_{\min}=4\sqrt{2}\Leftrightarrow x=y=z=\frac{\sqrt{2}}{3}\)
Lời giải:
Ta có:
A=√(x+y)(y+z)(z+x)(√y+zx+√z+xy+√x+yz)
A=(y+z)√(x+y)(x+z)x+(z+x)√(y+z)(y+x)y+(x+y)√(z+x)(z+y)z
Áp dụng BĐT Bunhiacopxky:
(x+y)(x+z)≥(x+√yz)2 và tương tự với những biểu thức khác suy ra:
A≥(y+z)(x+√yz)x+(z+x)(y+√xz)y+(x+y)(z+√xy)z
hay A≥2(x+y+z)+(y+z)√yzx+(z+x)√zxy+(x+y)√xyz
hay A≥2(x+y+z)+yz(y+z)√yz+xz(x+z)√xz+xy(x+y)√xyxyz M
Đặt (x,y,z)=(a2,b2,c2)
Khi đó: M=a3b3(a2+b2)+b3c3(b2+c2)+c3a3(a2+c2)a2b2c2
Áp dụng BĐT AM-GM:
a5b3+a3b5≥2√a8b8=2a4b4
b5c3+c5b3≥2b4c4
c5a3+a5c3≥2c4a4
⇒a3b3(a2+b2)+b3c3(b2+c2)+c3a3(c2+a2)≥2(a4b4+b4c4+c4a4) (1)
(cộng các BĐT theo vế)
Tiếp tục AM-GM:
a4b4+b4c4≥2a2b4c2;b4c4+c4a4≥2a2b2c4;c4a4+a4b4≥2a4b2c2
⇒a4b4+b4c4+c4a4≥a2b2c2(a2+b2+c2) (2)
Từ(1);(2)⇒a3b3(a2+b2)+b3c3(b2+c2)+c3a3(c2+a2)≥2a2b2c2(a2+b2+c2)
⇒M≥2(a2+b2+c2)=2(x+y+z)
Do đó: A≥2(x+y+z)+M≥4(x+y+z)⇔A≥4√2
Vậy Amin=4√2⇔x=y=z=√23
Bài 1:
\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )
Khi đó:
\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)
\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)
(Áp dụng BĐT Cauchy_Schwarz)
Theo BĐT Cauchy dễ thấy:
\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$
Bài 2:
Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)
Ta có:
\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)
\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)
Áp dụng BĐT Cauchy:
\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)
\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)
\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)
Nhân theo vế:
\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)
\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)
\(\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$
Ta có: \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge\dfrac{3x}{4}\)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\dfrac{6x-y-z-2}{8}\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\dfrac{6y-z-x-2}{8}\left(2\right)\\\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{6z-x-y-2}{8}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{6x-y-z-2}{8}+\dfrac{6y-z-x-2}{8}+\dfrac{6z-x-y-2}{8}\)
\(=\dfrac{1}{2}\left(x+y+z\right)-\dfrac{3}{4}\ge\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\)
\(Q\ge2\left(x+y+z\right)+3.\frac{9}{x+y+z}=2\left(x+y+z\right)+\frac{27}{x+y+z}.\)
Đặt X+Y+Z=t (\(t\le1\))
\(Q\ge2t+\frac{27}{t}=\left(2t+\frac{2}{t}\right)+\frac{25}{t}\ge2\sqrt{2t.\frac{2}{t}}+\frac{25}{1}=4+25=29\\ \)
Dấu = xảy ra khi x=y=z=1/3
Theo bđt cô si ta có : \(x+y+z\ge3\sqrt[3]{xyz}\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)
=> \(Q\ge6\sqrt[3]{xyz}+9\sqrt[3]{\frac{1}{xyz}}\ge2\sqrt{6\sqrt[3]{xyz}\cdot9\sqrt[3]{\frac{1}{xyz}}}=6\sqrt{6}\)
Dấu = xảy ra khi : \(6\sqrt[3]{xyz}=9\sqrt[3]{\frac{1}{xyz}}\) Giải ra ta đc : \(xyz=\frac{3}{2}\sqrt{\frac{3}{2}}\)