Ta sẽ chứng minh:

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

Thật vậy, bình phương 2 vế, BĐT tương đương:

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

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

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

\(\Leftrightarrow a^2y^2+b^2x^2-2abxy\ge0\)

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

Áp dụng:

\(VT=\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\)

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

2, a, \(a+\dfrac{1}{a}\ge2\)

\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)

\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)

\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)

vậy...................

Câu 1:

\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)

\(=\sqrt{4+5}=3\)

\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

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

\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)

2b)

Biến đổi tương đương:

\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\) (1)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\dfrac{a+2\sqrt{ab}+b}{4}\)

\(\Leftrightarrow2a+2b\ge a+2\sqrt{ab}+b\)

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

=> (1) đúng

Dấu "=" xảy ra khi a = b.

2c)

Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:

\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\) (đpcm)

Dấu "=" xảy ra khi a = b.

2d)

Áp dụng BĐT AM - GM, ta có:

\(\dfrac{a^2+2}{\sqrt{a^2+1}}=\dfrac{a^2+1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\dfrac{1}{\sqrt{a^2+1}}\ge2\) (đpcm)

Dấu "=" xảy ra khi a = 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)

ko biet