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Ta có:

\(\left(\sqrt{a}.\dfrac{\sqrt{a}}{\sqrt{4a+3bc}}+\sqrt{b}\dfrac{\sqrt{b}}{\sqrt{4b+3ac}}+\sqrt{c}\dfrac{\sqrt{c}}{\sqrt{4c+3ab}}\right)^2\le\left(a+b+c\right)\left(\dfrac{a}{4a+3bc}+\dfrac{b}{4b+3ac}+\dfrac{c}{4c+3ab}\right)\)

\(=2\left(\dfrac{a}{4a+3bc}+\dfrac{b}{4b+3ac}+\dfrac{c}{4c+3ab}\right)\)

Nên ta chỉ cần chứng minh:

\(\dfrac{a}{4a+3bc}+\dfrac{b}{4b+3ac}+\dfrac{c}{4c+3ab}\le\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{4a}{4a+3bc}+\dfrac{4b}{4b+3ac}+\dfrac{4c}{4c+3ab}\le2\)

\(\Leftrightarrow\dfrac{3bc}{4a+3bc}+\dfrac{3ac}{4b+3ac}+\dfrac{3ab}{4c+3ab}\ge1\)

\(\Leftrightarrow\dfrac{bc}{4a+3bc}+\dfrac{ac}{4b+3ac}+\dfrac{ab}{4c+3ab}\ge\dfrac{1}{3}\)

Thật vậy, ta có:

\(VT=\dfrac{\left(bc\right)^2}{4abc+3\left(bc\right)^2}+\dfrac{\left(ca\right)^2}{4abc+3\left(ac\right)^2}+\dfrac{\left(ab\right)^2}{4abc+3\left(ab\right)^2}\)

\(VT\ge\dfrac{\left(ab+bc+ca\right)^2}{3\left(ab\right)^2+3\left(bc\right)^2+3\left(ca\right)^2+12abc}=\dfrac{\left(ab+bc+ca\right)^2}{3\left(ab\right)^2+3\left(bc\right)^2+3\left(ca\right)^2+6abc\left(a+b+c\right)}\)

\(VT\ge\dfrac{\left(ab+bc+ca\right)^2}{3\left(ab+bc+ca\right)^2}=\dfrac{1}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=...\)

- Với \(ab=0\), vai trò như nhau, giả sử 

\(b=0\Rightarrow Q=\dfrac{a}{c}+\sqrt{\dfrac{2c}{a}}=\dfrac{a}{c}+\dfrac{1}{2}\sqrt{\dfrac{2c}{a}}+\dfrac{1}{2}\sqrt{\dfrac{2c}{a}}\ge3\sqrt[3]{\dfrac{1}{2}}\)

- Với \(ab>0\)

\(Q=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\sqrt{\dfrac{2c}{a+b}}\ge\dfrac{\left(a+b\right)^2}{2ab+c\left(a+b\right)}+\sqrt{\dfrac{2c}{a+b}}\)

\(\ge\dfrac{\left(a+b\right)^2}{\dfrac{\left(a+b\right)^2}{2}+c\left(a+b\right)}+\sqrt{\dfrac{2c}{a+b}}=\dfrac{2}{\dfrac{2c}{a+b}+1}+\sqrt{\dfrac{2c}{a+b}}\)

Đặt \(\sqrt{\dfrac{2c}{a+b}}=x>0\)

\(\Rightarrow Q\ge\dfrac{2}{x^2+1}+x=\dfrac{x^3+x+2}{x^2+1}=\dfrac{x^3-2x^2+x}{x^2+1}+2=\dfrac{x\left(x-1\right)^2}{x^2+1}+2\ge2\)

\(\Rightarrow Q_{min}=2\) khi \(x=\left\{0;1\right\}\Rightarrow\left[{}\begin{matrix}c=0;a=b\\a=b=c\end{matrix}\right.\)

Bài 2:

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)

\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:

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

Thiết lập các BĐT tương tự:

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

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)

Dấu "=" không xảy ra nên ta có ĐPCM

Lưu ý: lần sau đăng từng bài 1 thôi nhé !

1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)

TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)

\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)

Cộng vế với vế ta được:

\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)

Ai lm dc bai 3 chua

\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)

Làm tương tự, cộng lại và rút gọn

d đâu ra vậy bạn ?

Đặt \(A=\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\Rightarrow A^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\)

Áp dụng BĐT Bu - nhi - a - cốp - xki ta có :

\(A^2=\left(\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\right)^2\le\left(1^2+1^2+1^2\right)\left(4a+3+4b+3+4c+3\right)=3\left[4\left(a+b+c\right)+9\right]=3\left(12+9\right)=63\)

\(\Rightarrow A=\sqrt{4a+3}+\sqrt{4b+3}+\sqrt{4c+3}\le\sqrt{63}=3\sqrt{7}\)

Dấu \("="\) xảy ra khi \(a=b=c=1\)