Cho x,y,z > 0 CMR : \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{\left(a+b+c\right)}{3}\)
1. a) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Tìm max \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+zx+6}}\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=8\end{matrix}\right.\). Min \(P=\frac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
c) \(x,y,z>0.\) Min \(P=\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}+\sqrt{\frac{y^3}{y^3+\left(z+x\right)^3}}+\sqrt{\frac{z^3}{z^3+\left(x+y\right)^3}}\)
d) \(a,b,c>0;a^2+b^2+c^2+abc=4.Cmr:2a+b+c\le\frac{9}{2}\)
e) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{2}\)
f) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=4\end{matrix}\right.\) Cmr: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\)
g) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=2\end{matrix}\right.\) Max : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Nguyễn Ngọc Lộc , ?Amanda?, Phạm Lan Hương, Akai Haruma, @Trần Thanh Phương, @Nguyễn Việt Lâm,
@tth_new
Giúp em vs ạ! Thanks nhiều ạ
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
Cho a,b,c > 0. CMR: (a + b + c)2 \(\ge\) 3(ab + bc + ca)
và \(\frac{\left(a+b+c\right)^2}{ab+bc+ca}+\frac{ab+bc+ca}{\left(a+b+c\right)^2}\ge\frac{10}{3}\)
Bài 1:Cho a,b,c,d là các số dương. Chứng minh rằng :
\(\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}+\frac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\ge\frac{a+b+c+d}{4}\)
Bài 2:Cho \(a>0,b>0,c>0\).\(CM:\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bài 3: a) Cho x,y,>0. CMR:\(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\)
b) Chứng minh rằng\(\Sigma\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)
Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)
Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)
\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)
Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*
\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)
\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)
Đẳng thức xảy ra khi a = b = c
P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:
1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)
bài 2 xem có ghi nhầm ko
3a biến đổi tí là xong
b tuong tự bài 1
2. a) \(\left\{{}\begin{matrix}x,y,z>1\\x+y+z=xyz\end{matrix}\right.\) Tìm min \(P=\frac{x-1}{y^2}+\frac{y-1}{z^2}+\frac{z-1}{x^2}\)
b) \(a,b,c>0.Cmr:\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
c) \(\left\{{}\begin{matrix}x,y,z\ge0\\x^2+y^2+z^2=2\end{matrix}\right.\) Tìm max \(P=\frac{x^2}{x^2+yz+x+1}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}\)
d) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{ab+3c}+\frac{b}{bc+3a}+\frac{c}{ca+3b}\ge\frac{3}{4}\)
\(A=\frac{a}{ab+c\left(a+b+c\right)}+\frac{b}{bc+a\left(a+b+c\right)}+\frac{c}{ca+b\left(a+b+c\right)}\)
\(=\frac{a}{\left(b+c\right)\left(a+c\right)}+\frac{b}{\left(a+b\right)\left(a+c\right)}+\frac{c}{\left(a+b\right)\left(c+b\right)}\)
Áp dụng bđt AM-GM ta có
\(A=\frac{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\ge27.\frac{a^2+b^2+c^2+ab+bc+ca}{8\left(a+b+c\right)^3}\)\(=\frac{a^2+b^2+c^2+ab+bc+ca}{8}\)
\(=\frac{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}{8}\)\(\ge\frac{9-\frac{\left(a+b+c\right)^2}{3}}{8}=\frac{9-3}{8}=\frac{3}{4}\)
Dấu "=" xảy ra khi a=b=c=1
b) Mạnh hơn, và dễ dàng hơn là:
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{\sum c\left(a-b\right)^2}{abc}\)
Nó tương đương với: \({\frac {{a}^{2}}{{b}^{2}}}+{\frac {{b}^{2}}{{c}^{2}}}+{\frac {{c}^{2} }{{a}^{2}}}+3-2\,{\frac {a}{b}}-2\,{\frac {b}{c}}-2\,{\frac {c}{a}} \geqq 0\)
Là hiển nhiên vì \(\frac{a^2}{b^2}+1\ge\frac{2a}{b}\)
Đơn giản:))
a) Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow ab+bc+ca=1;0< a,b,c< 1\)
Cần chứng minh: \(P=\sum\frac{\frac{1}{a}-1}{\frac{1}{b^2}}=\sum\frac{b^2-ab^2}{a}\ge\sqrt{3}-1\)
Hay là: \(\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)\sqrt{ab+bc+ca}\ge\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)+a^2+b^2+c^2\)
\(\Leftrightarrow\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)^2\left(ab+bc+ca\right)\ge\) \(\Big[ (\sqrt{3} -1) (ab+bc+ca) +a^2+b^2+c^2\Big]^2\)
Giả sử \(c=\min\{a,b,c\}\) và đặt \(a=c+u, \, b=c+v \, (u,\, v \geq 0)\)
Nếu mình không nhìn nhầm, sau khi rút gọn, nhóm lại theo biến c, bạn nhận được một cái gì đó gọi là hiển nhiên
Chúc may mắn, mình mới rút gọn thử thì thấy có vẻ hiển nhiên thật :))
Bài 1 :Cho a,b,c dương thỏa mãn a+b+c=2
CMR \(\frac{bc}{\sqrt{3a^2+4}}+\frac{ca}{\sqrt{3b^2+4}}+\frac{ab}{\sqrt{3c^2+4}}\ge\frac{\sqrt{3}}{3}\)
Bài 2:Cho a,b,c>0. CMR
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
Cho a,b,c \(\ge\)0 .
CMR: \(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Cho a,b,c > 0. CMR:
1. \(a^3+b^3+c^3\ge3abc\)
2. \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)
3. \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)
4. \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
5. \(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}\ge\frac{1}{ab+1}\)
6.\(\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}\ge\frac{3}{1+abc}\)
Cho a;b;c>0.CMR:
\(\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}+\sqrt[3]{\frac{b^2+ca}{abc\left(c^2+a^2\right)}}+\sqrt[3]{\frac{c^2+ab}{abc\left(a^2+b^2\right)}}\ge\frac{9}{a+b+c}\)