Áp dụng BĐT Cauchy-Schwarz ta có:

\(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\)

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

\(=\sqrt{4\left(a+b\right)^2}=2\left(a+b\right)\)

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

Áp dụng Cauchy-Schwarz ta có:

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

Ta có: 

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

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

Dấu = xảy ra khi \(a=b\)

Áp dụng BĐT Cauchy-Schwarz ta có: 

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

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

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

Đẳng thức xảy ra khi \(a=b\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì \(x,y,z>0\)và ta cần chứng minh \(\frac{x}{\sqrt{3zx+yz}}+\frac{y}{\sqrt{3xy+zx}}+\frac{z}{\sqrt{3yz+xy}}\ge\frac{3}{2}\)\(\Leftrightarrow\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\frac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng phân thức, ta có: \(\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\)\(\frac{\left(x+y+z\right)^2}{x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}}\)

Áp dụng BĐT Cauchy-Schwarz, ta có: \(x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}\)\(=\sqrt{x}.\sqrt{3zx^2+xyz}+\sqrt{y}.\sqrt{3xy^2+xyz}+\sqrt{y}.\sqrt{3yz^2+xyz}\)\(\le\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\)

Ta cần chứng minh \(\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\le\frac{2}{3}\left(x+y+z\right)^2\)

\(\Leftrightarrow\left(x+y+z\right)^4\ge\frac{9}{4}\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]\)

\(\Leftrightarrow\left(x+y+z\right)^3\ge\frac{27}{4}\left(xy^2+yz^2+zx^2+xyz\right)\)(*)

Không mất tính tổng quát, giả sử \(y=mid\left\{x,y,z\right\}\)thì khi đó \(\left(y-x\right)\left(y-z\right)\le0\Leftrightarrow y^2+zx\le xy+yz\)

\(\Leftrightarrow xy^2+zx^2\le x^2y+xyz\Leftrightarrow xy^2+yz^2+zx^2+xyz\le\)\(x^2y+yz^2+2xyz=y\left(z+x\right)^2=4y.\frac{z+x}{2}.\frac{z+x}{2}\)

\(\le\frac{4}{27}\left(y+\frac{z+x}{2}+\frac{z+x}{2}\right)^3=\frac{4\left(x+y+z\right)^3}{27}\)

Như vậy (*) đúng

Đẳng thức xảy ra khi a = b = c

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

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

Mình làm phần dễ nhất rồi, còn lại của bạn đó ^^

Đặt . Do đó . Cần chứng minh:

Or 

Bình phương 2 vế và xét hiệu, ta cần chứng minh:

Đó là điều hiển nhiên vì: 

Done.

Áp dụng BĐT \(\sqrt{xy}\le\frac{x+y}{2}\)

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

\(\Rightarrow VT\ge\frac{2\left(a+b+c\right)}{\frac{4a+a+3b}{2}+\frac{4b+b+3c}{2}+\frac{4c+c+3a}{2}}\)

\(\Rightarrow VT\ge\frac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\frac{1}{2}\) (đpcm)

Dấu "=" khi \(a=b=c\)

Áp dụng BĐt bunhiakovsky ta có:

`(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})^2<=(a+b)(3a+b+3b+a)`

`<=>(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})^2<=4(a+b)^2`

`<=>\sqrt{a(3a+b)}+\sqrt{b(3b+a)}<=2(a+b)`

`=>(a+b)/(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})>=1/2`

Dấu "=" `<=>a=b`

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

Dấu "=" xảy ra khi và chỉ khi \(a=b\)

giải thích chỗ đang <= rồi chuyển sang >= là sao