Bài này hôm trước hình như bạn mới hỏi xong, vậy làm chi tiết cho đỡ băn khoăn:

Với các số dương a;b;c;x;y;z bất kì, ta chứng minh BĐT sau:

\(\sqrt{a^2+x^2}+\sqrt{b^2+y^2}\ge\sqrt{\left(a+b\right)^2+\left(x+y\right)^2}\)

Thật vậy, BĐT tương đương:

\(a^2+b^2+x^2+y^2+2\sqrt{a^2b^2+x^2y^2+x^2b^2+a^2y^2}\ge a^2+b^2+x^2+y^2+2ab+2xy\)

\(\Leftrightarrow\sqrt{a^2b^2+x^2y^2+a^2y^2+b^2x^2}\ge ab+xy\)

\(\Leftrightarrow a^2b^2+x^2y^2+a^2y^2+b^2x^2\ge a^2b^2+x^2y^2+2abxy\)

\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) (luôn đúng)

Từ đó suy ra:

\(\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\ge\sqrt{\left(a+b\right)^2+\left(x+y\right)^2}+\sqrt{c^2+z^2}\ge\sqrt{\left(a+b+c\right)^2+\left(x+y+z\right)^2}\)

Áp dụng cho bài toán:

\(VT=\sqrt{\left(x+\dfrac{y}{2}\right)^2+\left(\dfrac{\sqrt{3}y}{2}\right)^2}+\sqrt{\left(y+\dfrac{z}{2}\right)^2+\left(\dfrac{\sqrt{3}z}{2}\right)^2}+\sqrt{\left(z+\dfrac{x}{2}\right)^2+\left(\dfrac{\sqrt{3}x}{2}\right)^2}\)

\(VT\ge\sqrt{\left(x+\dfrac{y}{2}+y+\dfrac{z}{2}+z+\dfrac{x}{2}\right)^2+\left(\dfrac{\sqrt{3}y}{2}+\dfrac{\sqrt{3}z}{2}+\dfrac{\sqrt{3}x}{2}\right)^2}=2\left(x+y+z\right)\) (đpcm)

\(\Leftrightarrow\sqrt{4x^2+4xy+8y^2}+\sqrt{4y^2+4yz+8z^2}+\sqrt{4z^2+4zx+8x^2}\ge4\left(x+y+z\right)\)

Ta có:

\(VT=\sqrt{\left(2x+y\right)^2+\left(\sqrt{7}y\right)^2}+\sqrt{\left(2y+z\right)^2+\left(\sqrt{7}z\right)^2}+\sqrt{\left(2z+x\right)^2+\left(\sqrt{7}x\right)^2}\)

\(VT\ge\sqrt{\left(2x+y+2y+z+2z+x\right)^2+\left(\sqrt{7}x+\sqrt{7}y+\sqrt{7}z\right)^2}\)

\(VT\ge\sqrt{16\left(x+y+z\right)^2}=4\left(x+y+z\right)\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\)

Lời giải:
Ta thấy, với mọi $x,y,z$ là số thực thì:

$(x-y+z)^2\geq 0$

$\sqrt{y^4}\geq 0$

$|1-z^3|\geq 0$

$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|\geq 0$ với mọi $x,y,z$

Kết hợp $(x-y+z)^2+\sqrt{y^4}+|1-z^3|\leq 0$

$\Rightarrow (x-y+z)^2+\sqrt{y^4}+|1-z^3|=0$

Điều này xảy ra khi: $x-y+z=y^4=1-z^3=0$

$\Leftrightarrow y=0; z=1; x=-1$

a) \(A=2\sqrt{7}+3\sqrt{7}-2\sqrt{7}\\ A=3\sqrt{7}\)

b) \(\Leftrightarrow\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ x-y=x-y\)

a) \(A=2\sqrt{7}+3\sqrt{7}-2\sqrt{7}\\ A=3\sqrt{7}\)

b) \(\Leftrightarrow\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ x-y=x-y\)

đk : \(x\ge0;y\ge0;x\ne y\)

A = \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Leftrightarrow\) \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Leftrightarrow\) \(\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Leftrightarrow\) \(\dfrac{x-\sqrt{xy}-\sqrt{xy}-y}{x-y}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Rightarrow\) \(x-2\sqrt{xy}-y=2\sqrt{xy}\) \(\Leftrightarrow\) \(x-y=4\sqrt{xy}\)

\(\Leftrightarrow\) A = \(\dfrac{2\sqrt{xy}}{4\sqrt{xy}}=\dfrac{1}{2}\)

không biết sai chỗ nào ??? sao bài làm lại trái với câu hỏi thế này ???

\(A=\dfrac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\sqrt{x}+\sqrt{y}\\ A=\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}=2\sqrt{y}\)

Đề sai

\(A=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}+\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)

\(=\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}\)

\(=2\sqrt{x}\)

Ta có:

\(VT=\left(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{x}\cdot\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\right)\\ =\left(\frac{\sqrt{xy}\left[\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2\right]}{\sqrt{xy}}\right)\\ =x-y=VP\left(đpcm\right)\)

Vậy với x>0, y>0 ta có đpcm

\(\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)= x-y

=\(\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)

= \(x-y=x-y\)