Áp dụng liên tiếp AM-GM và Cauchy-Schwarz ta có:

\(\begin{align*} \dfrac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}&\ge \dfrac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left (a^2+b^2 \right )}}\\ &=\dfrac{a^2+ab+1}{\sqrt{a^2+ab+1}}\\ &=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\\ &=\dfrac{1}{\sqrt{5}}\sqrt{\left ( \dfrac{9}{4}+\dfrac{3}{4}+1+1 \right )\left [\left ( a+\dfrac{b}{2} \right )^2+\dfrac{3b^2}{4}+a^2+c^2 \right ]}\\ &\ge \dfrac{1}{\sqrt{5}}\left [ \dfrac{3}{2}\left (a+\dfrac{b}{2} \right )+\dfrac{3}{4}b+a+c \right ]\\ &=\dfrac{1}{\sqrt{5}}\left ( \dfrac{5}{2}a+\dfrac{3}{2}b+c \right ) \end{align*}\)

Chứng minh tương tự, cộng lại ta có đpcm.

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

bài này cuốn hút thật, lâu lắm ms thấy . xí bài này nhé nghĩ đã lát quay lại làm

Áp dụng liên tiếp AM - GM và Cauchy - Schwarz ta có :

\(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left(a^2+b^2\right)}}\)

   \(=\frac{a^2+ab+1}{\sqrt{a^2+ab+1}}\)

\(=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\)

\(=\frac{1}{\sqrt{5}}\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+a^2+c^2\right]}\)

\(\ge\frac{1}{\sqrt{5}}\left[\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{4}b+a+c\right]\)

\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)

Chứng minh tương tự và công lại ta có đpcm 

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

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 \(a=b=c=...\)

\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)

\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)

\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)