Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)và \(x+y\ge2.\sqrt{xy}\)( dấu ''='' xảy ra ở 2 bđt này khi x=y )

Ta có \(B=\frac{1}{a}+\frac{1}{b}+\frac{2}{a+b}\ge\frac{4}{a+b}+\frac{2}{a+b}=\frac{6}{a+b}\)

\(=\frac{6}{a+b}+\frac{3\left(a+b\right)}{2}-\frac{3.\left(a+b\right)}{2}\ge2\sqrt{\frac{6}{a+b}.\frac{3\left(a+b\right)}{2}}-\frac{3.2.\sqrt{ab}}{2}\)

\(=2\sqrt{9}-3.\sqrt{ab}=6-3=3\)

Dấu ''='' xảy ra khi \(\hept{\begin{cases}\frac{6}{a+b}=\frac{3.\left(a+b\right)}{2}\\a=b\\a.b=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{6}{2a}=\frac{3.2a}{2}\\a=b\\a.b=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}12a^2=12\\a=b\\a.b=1\end{cases}}\)\(\Leftrightarrow a=b=1\)

vì (a-1)² ≥ 0 nên a² +1 ≥ 2a  ∀mọi x    (1)

vì (b-1)² ≥ 0 nên b² +1 ≥ 2b ∀ mọi x      (2)

từ 1 và 2 ⇒ a²+b² ≥ 2a+2b

               ⇒ A≥ 2(a+b)=2

dấu''=' xảy ra khi a=b=1/2

Ta thấy \(ab\le\dfrac{a^2+b^2}{2}=1\) và \(a+b\le\sqrt{2\left(a^2+b^2\right)}=2\). Áp dụng BĐT B.C.S, ta được \(P=\dfrac{a^4}{ba^2+a^2}+\dfrac{b^4}{ab^2+b^2}\) \(\ge\dfrac{\left(a^2+b^2\right)^2}{ba^2+ab^2+a^2+b^2}=\dfrac{2^2}{ab\left(a+b\right)+2}\ge\dfrac{4}{1.2+2}=1\)

ĐTXR \(\Leftrightarrow a=b=1\)

Vậy GTNN của P là 1 khi \(a=b=1\)

Ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(\Rightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}\Leftrightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(a+b\right)^2\ge4ab\left(1\right)\\\left(a+b\right)^2\le2\left(a^2+b^2\right)\left(2\right)\end{cases}}\)

Theo đề bài:

\(a+b+3ab=1\)

\(\Leftrightarrow4\left(a+b\right)+12ab=4\)

\(\Leftrightarrow4\left(a+b\right)+3\left(a+b\right)^2\ge4\left(theo\left(1\right)\right)\)

\(\Leftrightarrow3\left(a+b\right)^2+4\left(a+b\right)-4\ge0\)

\(\Leftrightarrow\left(a+b+2\right)\left[3\left(a+b\right)-2\right]\ge0\)

\(\Leftrightarrow3\left(a+b\right)-2\ge0\left(a,b>0\Rightarrow a+b+2>0\right)\)

\(\Leftrightarrow a+b\ge\frac{2}{3}\)

`\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\ge\frac{4}{9}\left(theo\left(2\right)\right)\)

Áp dụng các kết quả trên, ta có:

\(\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2\le2\left(1-a^2+1-b^2\right)\)\(=4-2\left(a^2+b^2\right)\le4-\frac{4}{9}=\frac{32}{9}\)

\(\Rightarrow\sqrt{1-a^2}+\sqrt{1-b^2}\le\frac{4\sqrt{2}}{3}\)

Ta có: \(\frac{3ab}{a+b}=\frac{1-\left(a+b\right)}{a+b}=\frac{1}{a+b}-1\le\frac{1}{\frac{2}{3}}-1=\frac{1}{2}\)

\(\Rightarrow A\le\frac{4\sqrt{2}}{3}+\frac{1}{2}\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b+3ab=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\3a^2+2a-1=0\end{cases}\Leftrightarrow}a=b=\frac{1}{3}\left(a,b>0\right)}\)

Vậy max A là \(\frac{4\sqrt{2}}{3}+\frac{1}{2}\Leftrightarrow a=b=\frac{1}{3}\)

Áp dụng bđt : 1/a + 1/b >= 4/a+b thì :

p = 1/a + 1/b >= 4/a+b >= 4/\(2\sqrt{2}\)=  \(\sqrt{2}\)

Dấu "=" xảy ra <=> a=b=\(\sqrt{2}\)

Vậy ...............

Tk mk nha

Ta có : \(a^2+ab+b^2=\left(a+b\right)^2-ab\ge\left(a+b\right)^2-\frac{\left(a+b\right)^2}{4}=\frac{3\left(a+b\right)^2}{4}\)

\(\Rightarrow\sqrt{a^2+ab+b^2}\ge\frac{\sqrt{3}\left(a+b\right)}{2}\)

Tương tự : \(\sqrt{b^2+bc+c^2}\ge\frac{\sqrt{3}\left(b+c\right)}{2}\) ; \(\sqrt{c^2+ac+a^2}\ge\frac{\sqrt{3}\left(c+a\right)}{2}\)

Suy ra : \(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ac+a^2}\ge\frac{\sqrt{3}}{2}.2.\left(a+b+c\right)=\sqrt{3}\)

Vậy MIN B = \(\sqrt{3}\) \(\Leftrightarrow\begin{cases}a+b+c=1\\a=b=c\end{cases}\)

\(\Leftrightarrow a=b=c=\frac{1}{3}\)

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\dfrac{3}{2}\left(a^2+b^2\right)+\dfrac{1}{2}\left(a+b\right)^2}\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2+\dfrac{1}{2}\left(a+b\right)^2}=\dfrac{\sqrt{5}}{2}\left(a+b\right)\)

Tương tự:

\(\sqrt{2b^2+bc+2c^2}\ge\dfrac{\sqrt{5}}{2}\left(b+c\right)\) ; \(\sqrt{2c^2+ca+2a^2}\ge\dfrac{\sqrt{5}}{2}\left(c+a\right)\)

Cộng vế với vế:

\(P\ge\sqrt{5}\left(a+b+c\right)\ge\dfrac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3=\dfrac{\sqrt{5}}{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{9}\)