a) Bình phương 2 vế được: \(\frac{4ab}{a+b+2\sqrt{ab}}\le\sqrt{ab}\)

<=> \(4ab\le\sqrt{ab}\left(a+b\right)+2ab\)

<=>\(\sqrt{ab}\left(a+b\right)\ge2ab\)

<=>\(a+b\ge2\sqrt{ab}\)

<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)

Vậy \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\forall a,b>0\)

a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

Ta có: \(A-\frac{2}{3}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\)

\(=\frac{3\left(-5\sqrt{x}+2\right)}{3\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-17\sqrt{x}-51+51}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-17}{3}+\frac{17}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ

\(\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\forall x\) thỏa mãn ĐKXĐ

\(\Rightarrow\frac{17}{\sqrt{x}+3}-\frac{17}{3}\le\frac{17}{3}-\frac{17}{3}=0\forall x\) thỏa mãn ĐKXĐ

\(\Rightarrow A-\frac{2}{3}\le0\forall x\) thỏa mãn ĐKXĐ

nên \(A\le\frac{2}{3}\)(đpcm)

c) Ta có: \(C=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)

Vậy: Giá trị của C không phụ thuộc vào a,b(đpcm)

Lời giải:
Do $abc=1$ nên đặt:

\((\sqrt{a}, \sqrt{b}, \sqrt{c})=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})\) với $x,y,z>0$

Khi đó, bài toán trở thành: Cho $x,y,z>0$. CMR:

\(\frac{xz^2}{2z^2y+xy^2}+\frac{yx^2}{2x^2z+yz^2}+\frac{zy^2}{2y^2x+zx^2}\geq 1\)

Thật vậy, áp dụng BĐT Cauchy-Schwarz:

\(\frac{xz^2}{2z^2y+xy^2}+\frac{yx^2}{2x^2z+yz^2}+\frac{zy^2}{2y^2x+zx^2}=\frac{(xz)^2}{2xyz^2+(xy)^2}+\frac{(xy)^2}{2x^2yz+(yz)^2}+\frac{(yz)^2}{2xy^2z+(xz)^2}\)

\(\geq \frac{(xz+xy+yz)^2}{2xyz^2+(xy)^2+2x^2yz+(yz)^2+2xy^2z+(xz)^2}=\frac{(xy+yz+xz)^2}{(xy+yz+xz)^2}=1\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c=1$

thank you

Bài 2:

Ta có: \(a,b>0\) nên: \(\Rightarrow ab\le\frac{\left(a+b\right)^2}{4}\)

Lại có: \(\frac{x^3+8y^3}{x^3}=\left(1+\frac{2y}{x}\right)\left(1-\frac{2y}{x}+\frac{4y^2}{x^2}\right)\) \(\le\frac{\left(2x^2+4y^2\right)^2}{4x^4}\)

\(\Rightarrow\sqrt{\frac{x^3}{x^3+8y^3}}\ge\frac{2x^2}{2x^2+4y^2}\)

Tương tự như trên ta có được: \(\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\ge\frac{4y^2}{2y^2+\left(x+y\right)^2}\)

Lại có: \(\left(x+y\right)^2\le2\left(x^2+y^2\right)\) nên:

\(\Rightarrow2y^2+\left(x+y\right)^2\le2x^2+4y^2\)

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

\(\Rightarrow\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2x^2+4y^2}=1\)

\(\Rightarrow Min_P=1\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}4y^2\left(x-y\right)^2=0\\\left(x-y\right)^2\left(x^2+xy+2y^2\right)=0\end{matrix}\right.\Leftrightarrow x=y\)

`(1/(1+2))+(1/(1+2+3))+(1/(1+2+3+4))+...+(1/(1+2+3+..+99))`