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

\(P=\sum\frac{a}{\sqrt{\left(2a\right)^2+\left(b+c\right)^2}}\le\sqrt{2}\sum\frac{a}{2a+b+c}=\sqrt{2}\sum a\left(\frac{1}{a+b+a+c}\right)\le\frac{\sqrt{2}}{4}\sum\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3\sqrt{2}}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)

Trừ mỗi vế cho 1, ta có:

\(\frac{b-16a+16c}{4a}=\frac{c-16b+16a}{4b}=\frac{a-16c+16b}{4c}=\frac{a+b+c}{4.\left(a+b+c\right)}=\frac{1}{4}\)(vì a,b,c > 0 nên a+b+c>0)

\(\Leftrightarrow\hept{\begin{cases}b+16c=17a\\c+16a=17b\\a+16b=17c\end{cases}}\Leftrightarrow a=b=c\)

tự thay vào

\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{8}+\frac{c+1}{8}\ge\frac{3}{4}a\)\(\Leftrightarrow\)\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{3}{4}a-\frac{1}{8}b-\frac{1}{8}-\frac{1}{4}\)

\(\Sigma\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{1}{2}\left(a+b+c\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\) :) 

tích đi rồi ta làm

tích đi bạn

k mk nha

k mk nha!

#meo#

$A=\frac{64abc}{(a+b)(b+c)(c+a)}+1+\frac{16ab}{(b+c)(c+a)}+\frac{16bc}{(b+a)(c+a)}+\frac{16ac}{(a+b)(a+c)}+4.(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c})=4.(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c})+\frac{64abc}{(a+b)(b+c)(c+a)}+\frac{16ab(a+b)+16bc(b+c)+16ac(a+c)}{(a+b)(b+c)(c+a)}+1=4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})+\frac{64abc}{(a+b)(b+c)(c+a)}+\frac{16(a+b)(b+c)(c+a)-32abc}{(a+b)(b+c)(c+a)}+1=4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})+\frac{32abc}{(a+b)(b+c)(c+a)}+17=4\left [\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b}+\frac{4abc}{(a+b)(b+c)(c+a)} \right ]+\frac{16abc}{(a+b)(b+c)(c+a)}+17\geq 4.2+17+\frac{16abc}{(a+b)(b+c)(c+a)}=25+\frac{16abc}{(a+b)(b+c)(c+a)}> 25$

( Do áp dụng bđt Schur mở rộng là :$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{4abc}{(a+b)(b+c)(c+a)}\geq 2$