Đáp án B.

Đặt log₆ x = log₉ y = log₄ (2x + 2y) = t

Đáp án A

Ta có

Khi đó

\(VT=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)

\(VT\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{25}{2}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)

<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)

<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)

đặt x+y+z = t 

=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)

=> x=y=z 

ta lại có 

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

vì x=y=z  => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

gật gật

Đặt \(\dfrac{x}{z}=a;\dfrac{y}{z}=b\).

Theo gt ta có \(a+b\le1\).

BĐT cần chứng minh tương đương:

\(a^2+b^2+\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{21}{2}\).

Theo bđt AM - GM: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2;a^2+\dfrac{1}{16}a^2\ge\dfrac{1}{2};b^2+\dfrac{1}{16}b^2\ge\dfrac{1}{2};\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\ge\dfrac{15}{32}.\left(\dfrac{4}{a+b}\right)^2\ge\dfrac{15}{2}\).

Cộng vế với vế của các bđt trên lại ta có đpcm.

\(4\le\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\le\dfrac{1}{4}\left(\sqrt{x}+\sqrt{y}+2\right)^2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+2\ge4\)

\(\Rightarrow2\le\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\Rightarrow x+y\ge2\)

\(\Rightarrow P\ge\dfrac{\left(x+y\right)^2}{x+y}=x+y\ge2\)

Dấu "=" xảy ra khi \(x=y=1\)

Dạ có thể diễn đạt theo cách dễ hiểu cho đứa ngu lâu dốt bền như em được không ạ ? ._.

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$\text{VT}(1^2+1^2+1^2)\geq (1+\frac{x}{y+z}+1+\frac{y}{x+z}+1+\frac{z}{x+y})^2$

$\Leftrightarrow 3\text{VT}\geq (3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})^2$

$ = \left[3+\frac{x^2}{xy+xz}+\frac{y^2}{yz+yx}+\frac{z^2}{zy+zx}\right]^2$

$\geq \left[3+\frac{(x+y+z)^2}{2(xy+yz+xz)}\right]^2$

$\geq \left[3+\frac{3(xy+yz+xz)}{2(xy+yz+xz)}\right]^2=\frac{81}{4}$

$\Rightarrow \text{VT}\geq \frac{27}{4}$

Dấu "=" xảy ra khi $x=y=z>0$

Áp dụng BĐT Bunhiacopxky:

$\text{VT}(1^2+1^2+1^2)\ge (1+\genfrac{}{}{0ex}{}{x}{y+z}+1+\genfrac{}{}{0ex}{}{y}{x+z}+1+\genfrac{}{}{0ex}{}{z}{x+y}{)}^2$

$\iff 3\text{VT}\ge (3+\genfrac{}{}{0ex}{}{x}{y+z}+\genfrac{}{}{0ex}{}{y}{x+z}+\genfrac{}{}{0ex}{}{z}{x+y}{)}^2$

$={\left[3+\genfrac{}{}{0ex}{}{x^2}{xy+xz}+\genfrac{}{}{0ex}{}{y^2}{yz+yx}+\genfrac{}{}{0ex}{}{z^2}{zy+zx}\right]}^2$

$\ge {\left[3+\genfrac{}{}{0ex}{}{(x+y+z{)}^2}{2(xy+yz+xz)}\right]}^2$

$\ge {\left[3+\genfrac{}{}{0ex}{}{3(xy+yz+xz)}{2(xy+yz+xz)}\right]}^2=\genfrac{}{}{0ex}{}{81}{4}$

$\Rightarrow \text{VT}\ge \genfrac{}{}{0ex}{}{27}{4}$

Dấu "=" xảy ra khi $x=y=z>0$

Lời giải:

Áp dụng TCDTSBN:

$\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1$

$\Rightarrow x=y; y=z; z=x\Rightarrow x=y=z$

Khi đó:

$|x+y|=|z-1|$

$\Leftrightarrow |2x|=|x-1|$

$\Rightarrow 2x=x-1$ hoặc $2x=-(x-1)$

$\Rightarrow x=-1$ hoặc $x=\frac{1}{3}$ (đều thỏa mãn)

Vậy $(x,y,z)=(-1,-1,-1)$ hoặc $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)