Sửa \(\dfrac{1}{3}\rightarrow3\)

Từ \(a+b+c+ab+bc+ca=6abc\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)

Ta có: \(\dfrac{1}{a^2}+1\ge\dfrac{2}{a};\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)

Và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab};\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ac}\)

Cộng theo vế các BĐT trên ta có:

\(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\Leftrightarrow3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge12\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\ge4\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)

\("="\Leftrightarrow a=b=c=1\)

để sau nha giờ bận .-.

Cauchy-Schwarz dạng ENgel:

\(P=\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\)

\(=\dfrac{1}{2}\cdot3-\left(\dfrac{b}{4a+2b}+\dfrac{c}{4b+2c}+\dfrac{a}{4c+2a}\right)\)

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

\(\le\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}\)

\(=\dfrac{3}{2}-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=1\)

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

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

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

Thật vậy, ta có:

\(VT=\frac{b^2c^2}{4abc+3b^2c^2}+\frac{a^2c^2}{4abc+3a^2c^2}+\frac{a^2b^2}{4abc+3a^2b^2}\)

\(VT\ge\frac{\left(ab+bc+ca\right)^2}{3\left(a^2b^2+b^2c^2+c^2a^2\right)+12abc}=\frac{a^2b^2+b^2c^2+c^2a^2+2\left(a+b+c\right)abc}{3\left(a^2b^2+b^2c^2+c^2a^2+4abc\right)}\)

\(VT\ge\frac{a^2b^2+b^2c^2+c^2a^2+4abc}{3\left(a^2b^2+b^2c^2+c^2a^2+4abc\right)}=\frac{1}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)

\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{4c}{a+b}\)

\(A>\frac{2a}{a+b+c}+\frac{2b}{c+a+b}+\frac{2c}{a+b+c}\)

\(A>\frac{2\left(a+b+c\right)}{a+b+c}>2\Rightarrow dpcm\)

\(\Leftrightarrow\frac{9}{4a^2+b^2+c^2}+\frac{9}{a^2+4b^2+c^2}+\frac{9}{a^2+b^2+4c^2}\le\frac{9}{2}\)

Thật vậy, ta có:

\(\frac{9}{4a^2+b^2+c^2}=\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\)

Tương tự: \(\frac{9}{a^2+4b^2+c^2}\le\frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{b^2+c^2}\) ; \(\frac{9}{a^2+b^2+4c^2}\le\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{2c^2}\)

Cộng vế với vế:

\(VT\le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{b^2+c^2}+\frac{a^2}{a^2+c^2}+\frac{c^2}{a^2+c^2}=\frac{3}{2}+3=\frac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

ĐÂY MÀ LÀ toán 5 ạ??

Gọi A là vế trái của BĐT cần chứng minh. Không mất tính tổng quát, ta giả sử a + b + c = 3. Áp dụng BĐT AM - GM ta có:

\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{8bc\left(4a+4b+c\right)}}+\frac{ab\left(4a+4b+c\right)}{27}\)\(\ge\frac{1}{2}\left(a+b\right)\)

Suy ra 

             \(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}\)\(+\frac{ab\left(4a+4b+c\right)}{54}\ge\frac{1}{4}\left(a+b\right)\)

Tương tự

            \(\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\frac{bc\left(4b+4c+a\right)}{54}\ge\frac{1}{4}\left(b+c\right)\)

và       \(\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}+\frac{ca\left(4c+4a+b\right)}{54}\ge\frac{1}{4}\left(c+a\right)\)

Cộng ba BĐT trên ta có: 

           \(\frac{1}{2\sqrt{2}}A\ge B\)

Với \(A=\frac{1}{54}[ab\left(4a+4b+c\right)+bc\left(4b+4c+a\right)\)

\(+ca\left(4c+4a+b\right)]\)

\(=\frac{1}{54}\left[4ab\left(a+b\right)+4bc\left(b+c\right)+4ca\left(c+a\right)+3abc\right]\)

\(=\frac{1}{54}\left[4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\right]\)

\(\le\frac{1}{54}\left(a+b+c\right)^3=\frac{1}{2}\)

và \(B=\frac{1}{4}.2\left(a+b+c\right)=\frac{3}{2}\)

Suy ra \(\frac{1}{2\sqrt{2}}A\ge\frac{3}{2}-\frac{1}{2}=1\Rightarrow A\ge2\sqrt{2}\)

Vậy 

              \(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{bc\left(4a+4b+c\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4a+b\right)}}\ge2\sqrt{2}\)(đpcm)

toán lớp 5 phiên bản hack não

Lời giải:

Để ý rằng:

\(\frac{4a^2+(b-c)^2}{2a^2+b^2+c^2}=\frac{2(2a^2+b^2+c^2)-2(b^2+c^2)+(b-c)^2}{2a^2+b^2+c^2}=2-\frac{(b+c)^2}{2a^2+b^2+c^2}\)

Biến đổi tương tự với các phân thức còn lại:

\(\Rightarrow \text{VT}=6-\underbrace{\left[\frac{(b+c)^2}{2a^2+b^2+c^2}+\frac{(c+a)^2}{2b^2+a^2+c^2}+\frac{(a+b)^2}{2c^2+a^2+b^2}\right]}_{N}\)

Ta muốn CM \(\text{VT}\geq 3\Leftrightarrow N\leq 3\) . Thật vậy:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{(b+c)^2}{2a^2+b^2+c^2}\leq \frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\). Tương tự như vậy:

\(\left\{\begin{matrix} \frac{(a+c)^2}{2b^2+a^2+c^2}\leq \frac{a^2}{a^2+b^2}+\frac{c^2}{c^2+b^2}\\ \frac{(a+b)^2}{2c^2+a^2+b^2}\leq \frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\end{matrix}\right.\)

Cộng theo vế thu được \(N\leq \frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}=3\)

CM hoàn tất. Dấu bằng xảy ra khi \(a=b=c>0\)