Cho a, b,c,d dương.
Chứng minh rằng \(\frac{2a}{a^6+b^4}+\frac{2b}{b^6+c^4}+\frac{2c}{c^6+a^4}\le\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\)
\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)
cho các số thực dương a,b,c chứng minh rằng
\(\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\le\frac{1}{4}\left(a+b+c\right)\)
\(\frac{ab}{a+b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Làm tương tự với 2 phân thức còn lại rồi cộng vào ra đpcm
Cho các số thực \(a,b,c\ge1\). Chứng minh rằng:
\(\frac{1}{2a-1}+\frac{1}{2b-1}+\frac{1}{2c-1}+3\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)
\(\frac{1}{2a-1}+\frac{1}{1}\ge\frac{4}{2a}=\frac{2}{a}\) ; \(\frac{1}{2b-1}+\frac{1}{1}\ge\frac{2}{b}\) ; \(\frac{1}{2c-1}+\frac{1}{1}\ge\frac{2}{c}\)
\(\Rightarrow VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(\Rightarrow VT\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho các số thực \(a,b,c>1\)
Chứng minh rằng:\(\frac{1}{2a-1}+\frac{1}{2b-1}+\frac{1}{2c-1}+3\ge\frac{4}{a-b}+\frac{4}{b-c}+\frac{4}{c-a}\)
Sửa đề:
Cho a, b, c > 1(chỗ này là ý tui, dùng Wolfram Alpha sẽ thấy nếu không sửa như vầy thì đẳng thức không xảy ra). CMR:
\(\frac{1}{2a-1}+\frac{1}{2b-1}+\frac{1}{2c-1}+3\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\) (cái này là ý chủ tus đấy nhá!)
\(\Leftrightarrow\frac{2a}{2a-1}+\frac{2b}{2b-1}+\frac{2c}{2c-1}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\) (tách ghép vế trái + làm chặt BĐT do \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b};..\))
\(\Leftrightarrow\frac{2a^2-4a+2}{a\left(2a-1\right)}+\frac{2b^2-4b+2}{b\left(2b-1\right)}+\frac{2c^2-4c+1}{c\left(2c-1\right)}\ge0\) (chuyển vế + quy đồng)
\(\Leftrightarrow\frac{2\left(a-1\right)^2}{a\left(2a-1\right)}+\frac{2\left(b-1\right)^2}{b\left(2b-1\right)}+\frac{2\left(c-1\right)^2}{c\left(2c-1\right)}\ge0\) (đúng)
Đẳng thức xảy ra khi a = b = c = 1
Vậy ta có đpcm.
\(\frac{1}{2a-1}+1\ge\frac{\left(1+1\right)^2}{2a-1+1}=\frac{4}{2a}=\frac{2}{a}\)
HSG Bắc Ninh 2018-2019
Có \(\frac{1}{2a-1}\ge\frac{1}{a^2}\);\(\frac{1}{2b-1}\ge\frac{1}{b^2}\);\(\frac{1}{2c-1}\ge\frac{1}{c^2}\)
\(\Rightarrow\text{Σ}_{cyc}\frac{1}{2a-1}+3\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+3\)
Lại có \(\frac{1}{a^2}+\frac{1}{b^2}\ge^{co-si}\frac{2}{ab}\ge\frac{8}{\left(a+b\right)^2}\)
\(\frac{8}{\left(a+b\right)^2}+2\ge\frac{8}{a+b}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+2\ge\frac{8}{a+b}\)
Tương tự ta có \(\frac{1}{b^2}+\frac{1}{c^2}+2\ge\frac{8}{b+c}\);\(\frac{1}{c^2}+\frac{1}{a^2}+2\ge\frac{8}{c+a}\)
\(\Rightarrow2\left(\text{Σ}_{cyc}\frac{1}{a^2}+3\right)\ge\text{Σ}_{cyc}\frac{8}{a+b}\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+3\ge\text{Σ}_{cyc}\frac{4}{a+b}\)
\(\RightarrowĐPCM\left("="\Leftrightarrow a=b=c=1\right)\)
Cho a,b,c là các số thực dương thỏa mãn a4b4+b4c4+c4a4=3a4b4c4.
Chứng minh rằng:\(\frac{1}{a^3b+2c^2+1}+\frac{1}{b^3c+2a^2+1}+\frac{1}{c^3a+2b^2+1}\le\frac{3}{4}\)
\(GT\Rightarrow\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}=3\)
Ta có: \(\frac{1}{a^4}+\frac{1}{a^4}+\frac{1}{a^4}+\frac{1}{b^4}\ge4\sqrt[4]{\frac{1}{a^{12}b^4}}=\frac{4}{a^3b}\)
Tương tự: \(\frac{3}{b^4}+\frac{1}{c^4}\ge\frac{4}{b^3c}\) ; \(\frac{3}{c^4}+\frac{1}{a^4}\ge\frac{4}{c^3a}\)
\(\Rightarrow\frac{1}{a^3b}+\frac{1}{b^3c}+\frac{1}{c^3a}\le\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}=3\)
\(VT=\frac{1}{a^3b+c^2+c^2+1}+\frac{1}{b^3c+a^2+a^2+1}+\frac{1}{c^3a+b^2+b^2+1}\)
\(VT\le\frac{1}{16}\left(\frac{1}{a^3b}+\frac{2}{c^2}+1+\frac{1}{b^3c}+\frac{2}{a^2}+1+\frac{1}{c^3a}+\frac{2}{b^2}+1\right)\)
\(VT\le\frac{1}{16}\left(\frac{1}{a^3b}+\frac{1}{b^3c}+\frac{1}{c^3a}+2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+3\right)\)
\(VT\le\frac{1}{16}\left(6+2\sqrt{3\left(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right)}\right)=\frac{1}{16}\left(6+6\right)=\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Chứng minh rằng với mọi a, b, c > 0 ta có: \(\frac{a^4}{1+a^2b}+\frac{b^4}{1+b^2c}+\frac{c^4}{1+c^2a}\ge\frac{abc\left(a+b+c\right)}{1+abc}\)
Cho a, b, c thỏa mãn \(\frac{a}{2a+b+c}+\frac{b}{2b+c+a}+\frac{c}{2c+a+b}=\frac{3}{4}.\)
Chứng minh rằng \(\frac{a^2}{2a+b+c}+\frac{b^2}{2b+c+a}+\frac{c^2}{2c+a+b}=\frac{a+b+c}{4}.\)
cho a,b,c,d >0 thỏa a+b+c+d=4 chứng minh \(\frac{a}{1+b^2c}+\frac{b}{1+c^2a}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\)
\frac{a}{1+b^{2}c}+\frac{b}{1+c^{2}d}+\frac{c}{1+d^{2}a}+\frac{d}{1+a^{2}b}\geq 2$
Ta có $\sum \frac{a}{1+b^2c}=\sum \frac{a^2}{a+ab^2c}$
Áp dụng Cauchy-Schwarzt ta có
$\sum \frac{a}{1+b^2c}=\sum \frac{a^2}{a+ab^2c}\geq \frac{(a+b+c+d)^2}{a+b+c+d+ab^2c+bc^2d+cd^2a+da^2b}=\frac{16}{4+ab^2c+bc^2d+cd^2a+da^2b}$
Do đó ta chỉ cần chứng minh $ab^2c+bc^2d+cd^2a+da^2b\leq 4$ là suy ra $\sum \frac{a}{1+b^2c}\geq \frac{16}{4+4}=2$
Bất đẳng thức đã cho tương đương $ab.bc+bc.cd+cd.da+da.ab\leq 4$ với $a+b+c+d=4$
Chuyển $\left ( ab,bc,cd,da \right )\Rightarrow (x,y,z,t)$
Ta có $x+y+z+t=ab+bc+cd+ad \leq \frac{(a+b+c+d)^2}{4}=4$
Lại có $ab^2c+bc^2d+cd^2a+da^2b=xy+yz+zt+tx \leq \frac{(x+y+z+t)^2}{4} \leq \frac{4^2}{4}=4$
Vậy ta có đpcm
Dấu = xảy ra khi $a=b=c=d=1$
doc lam sao