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Lời giải:

Sử dụng PP khai triển :

\(\frac{a+b}{\sqrt{a(3a+b)+b(3b+a)}}\geq \frac{1}{2}\)

\(\Leftrightarrow \frac{(a+b)^2}{a(3a+b)+b(3b+a)}\geq \frac{1}{4}\)

\(\Leftrightarrow 4(a+b)^2\geq a(3a+b)+b(3b+a)\)

\(\Leftrightarrow a^2+b^2+6ab\geq 0\)

\(\Leftrightarrow (a+b)^2+4ab\geq 0\). Điều này luôn đúng với \(a,b\geq 0\) tuy nhiên dấu bằng không xảy ra do \(a,b\neq 0\)

Do đó: \(\frac{a+b}{\sqrt{a(3a+b)+b(3b+a)}}> \frac{1}{2}\)

mk nghĩ đề bài như này ms đúng chứ

\(\dfrac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\dfrac{1}{2}\)

vs a,b>0

cm \(vt=\dfrac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\)

\(\ge\dfrac{2\left(a+b\right)}{\dfrac{4a+3a+b}{2}+\dfrac{4b+3b+a}{2}}=\dfrac{2\left(a+b\right)}{\dfrac{8\left(a+b\right)}{2}}=\dfrac{1}{2}\)(dpcm)

dau = xay ra khi a=b>0

các bạn ơi giúp mình với

Áp dụng BĐT AM-GM: \(\dfrac{1}{2}\sqrt{\left(a+3b\right)\left(b+3a\right)}\le\dfrac{1}{4}\left(4a+4b\right)=a+b\)

Ta chứng minh: \(3\left(a+b\right)^2+4ab\ge2\left(a+b\right)\)

hay \(3\left(a+b\right)^2+4ab\ge2\left(a+b\right)\left(\sqrt{a}+\sqrt{b}\right)^2\)

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

Dấu = xảy ra khi \(a=b=\dfrac{1}{4}\)

+) Ta có \(\sqrt{4a\left(3a+b\right)}\le\frac{4a+\left(3a+b\right)}{2}=\frac{7a+b}{2}\)

\(\Rightarrow\sqrt{a\left(3a+b\right)}\le\frac{7a+b}{4}\left(2\right)\)

+) Tương tự ta lại có :

\(\sqrt{b\left(3b+a\right)}\le\frac{7b+a}{4}\left(3\right)\)

+) Từ (2) và (3) ta có :

\(VT\left(1\right)\ge\frac{a+b}{\frac{7a+b}{4}+\frac{7b+a}{4}}=\frac{1}{2}\left(đpcm\right)\)

Ta có: \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\)

\(=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{2\left(a+b\right)}{\frac{1}{2}\left(4a+3a+b\right)+\frac{1}{2}\left(4b+3b+a\right)}\) (Cauchy)

\(=\frac{2\left(a+b\right)}{4\left(a+b\right)}=\frac{1}{2}\)

Dấu "=" xảy ra khi: a = b

Áp dụng BĐT AM-GM ta có: 

\(2\sqrt{a\left(3a+b\right)}=\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}=\frac{7a+b}{2}\)

\(2\sqrt{b\left(3b+a\right)}=\sqrt{4b\left(3b+a\right)}\le\frac{4b+3b+a}{2}=\frac{7b+a}{2}\)

Suy ra \(\sqrt{b\left(3b+a\right)}+\sqrt{a\left(3a+b\right)}\le\frac{8a+8b}{4}=2\left(a+b\right)\)

\(\Rightarrow\frac{a+b}{\sqrt{b\left(3b+a\right)}+\sqrt{a\left(3a+b\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)

\(\frac{4\left(a+b\right)}{2\sqrt{4a\left(3a+b\right)}+2\sqrt{4b\left(3b+a\right)}}\ge\frac{4\left(a+b\right)}{4a+3a+b+4b+3b+a}=\frac{4\left(a+b\right)}{8\left(a+b\right)}=\frac{1}{2}\)

Dấu "=" xảy ra khi \(a=b\)

\(\sqrt{a\left(8a+b\right)}=\dfrac{1}{3}\cdot\sqrt{9a\left(8a+b\right)}< =\dfrac{1}{3}\cdot\dfrac{9a+8a+b}{2}=\dfrac{1}{6}\left(17a+b\right)\)

\(\sqrt{b\left(8b+a\right)}< =\dfrac{1}{6}\left(17b+a\right)\)

=>\(\sqrt{a\left(8a+b\right)}+\sqrt{b\left(8b+a\right)}< =3\left(a+b\right)\)

=>\(\dfrac{a+b}{\sqrt{a\left(8a+b\right)}+\sqrt{b\left(8b+a\right)}}>=\dfrac{1}{3}\)