1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)

2) Cho a,b,c>0 tm a^2+b^2+c^2 bé hơn hoặc bằng abc. Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)

3) Cho a,b,c>0 tm a+b+c<=3. Cmr \(\frac{ab}{\sqrt{3+c}}+\frac{bc}{\sqrt{3+a}}+\frac{ca}{\sqrt{3+b}}\le\frac{3}{2}\)

4) Cho a,b,c>0 tm a+b+c=2. Cmr \(\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le1\)

5) Cho a,b,c>0. Cmr \(\sqrt{\frac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\frac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\frac{c^3}{5c^2+\left(a+b\right)^2}}\le\sqrt{\frac{a+b+c}{3}}\)

6) Cho a,b,c>0. Cmr \(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\frac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\frac{c^2}{\left(2c+a\right)\left(2c+b\right)}\le\frac{1}{3}\)

Giúp mình với nhé các bạn

1. cho \(0< a\le b\le c\) . Cmr: \(\frac{2a^2}{b^2+c^2}+\frac{2b^2}{c^2+a^2}+\frac{2c^2}{a^2+b^2}\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)

2. cho \(a,b,c\ge0\). cmr: \(a^2+b^2+c^2+3\sqrt[3]{\left(abc\right)^2}\ge2\left(ab+bc+ca\right)\)

3. \(a,b,c>0.\) Cmr: \(\sqrt{\left(a^2b+b^2c+c^2a\right)\left(ab^2+bc^2+ca^2\right)}\ge abc+\sqrt[3]{\left(a^3+abc\right)\left(b^3+abc\right)\left(c^3+abc\right)}\)

4. \(a,b,c>0\). Tìm Min \(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4\)

CHO a,b,c>0 thỏa mãn: \(a^2b^2+b^2c^2+c^2a^2\ge a^2+b^2+c^2\)

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

ĐẶT \(A=\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(c^2+a^2\right)}\)

ĐẶT:\(\frac{1}{a}=x,\frac{1}{y}=b,\frac{1}{z}=c\)

\(\Rightarrow x^2+y^2+z^2\ge1\)

\(\Rightarrow A=\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{z^2+y^2}\)

TA CÓ:

\(x\left(y^2+z^2\right)=\frac{1}{\sqrt{2}}\sqrt{2x^2\left(y^2+z^2\right)\left(y^2+z^2\right)}\le\frac{1}{\sqrt{2}}\sqrt{\frac{\left(2x^2+2y^2+2z^2\right)^3}{27}}=\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)TƯƠNG TỰ:

\(y\left(x^2+z^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2},z\left(x^2+y^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)LẠI CÓ:
\(A=\frac{x^3}{y^2+z^2}+\frac{y^3}{x^2+z^2}+\frac{z^3}{x^2+y^2}=\frac{x^4}{x\left(y^2+z^2\right)}+\frac{y^4}{y\left(x^2+z^2\right)}+\frac{z^4}{z\left(x^2+y^2\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)}\ge\frac{1}{3.\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}} \)\(\ge\frac{\sqrt{3}}{2}\sqrt{x^2+y^2+z^2}\ge\frac{\sqrt{3}}{2}\)

DẤU BẰNG XẢY RA\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\Rightarrow DPCM\)

\(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)

\(VP=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\le\frac{a+b}{2}\sqrt{2\left(a+b\right)}\)\(\Rightarrow\)\(VP^2\le\frac{\left(a+b\right)^3}{2}\) (1) 

chứng minh bổ đề: \(VT^2=\left(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\right)^2\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\frac{\left(a+b\right)^4}{4}+\frac{\left(a+b\right)^2}{16}+\frac{\left(a+b\right)^3}{4}\ge\frac{\left(a+b\right)^3}{2}\)

\(\Leftrightarrow\)\(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge\left(a+b\right)^3\)

Có: \(\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge2\sqrt{\frac{\left(a+b\right)^6}{4}}=\left(a+b\right)^3\)\(\Rightarrow\)\(VT^2\ge\frac{\left(a+b\right)^3}{2}\) (2) 

(1) và (2) => \(VT^2\ge VP^2\) => \(VT\ge VP\) ( đpcm ) 

Câu 2:

Tìm GTLN của biểu thức sau :

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

Giải

ĐK: \(\frac{5}{3}\le x\le\frac{7}{3}\)

Cmr: \(a+b\le2\sqrt{ab}\left(a,b\ge0\right)\)(*)

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) luôn đúng với a,b>0

Dấu "=" xảy ra <=> a=b

Ta có \(A=\sqrt{3x-5}+\sqrt{7-3x}\)

=> \(A^2=\left[3x-5+7-3x+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\right]=2+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)

Áp dụng BĐT (*) ta được:

\(A^2\le2+\left(3x-5\right)+\left(7-3x\right)=4̸\)

\(\Rightarrow A\le2\)

Vậy MaxA=2 <=> \(\left\{{}\begin{matrix}\frac{5}{3}\le x\le\frac{7}{3}\\3x-5=7-3x\end{matrix}\right.\)=>x=2

\(ab+bc+ca=1\)\(\Rightarrow\)\(\hept{\begin{cases}a+b+c\ge\sqrt{3}\\a^2+b^2+c^2\ge1\end{cases}}\)

\(\left(a-\frac{1}{\sqrt{3}}\right)^2\ge0\)\(\Leftrightarrow\)\(a\le\frac{\sqrt{3}}{2}a^2+\frac{\sqrt{3}}{6}\)

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

dễ cm \(\frac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{4\left(ab+bc+ca\right)}\ge2\left(a+b+c\right)\)

\(2\sqrt{a^2-ab+b^2}=2\sqrt{\left(\frac{a^2}{b}-a+b\right)b}\le a^2-a+2b\)

từ đó bđt cần cm <=> \(a+b+c\ge ab+bc+ca\)

lại có \(ab+bc+ca+abc\le4\)

\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)\le\left(a+2\right)\left(b+2\right)+...\)

\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\ge1\)

\(\Leftrightarrow\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\le1\)

\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\le\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\le1\)

\(\Rightarrow a+b+c\ge ab+bc+ca\)

=>Q.E.D

1, cho a,b,c là các số thực dương chứng minh rằng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2a+b}{a\left(a+2b\right)}+\frac{2b+c}{b\left(b+2c\right)}+\frac{2c+a}{c\left(a+2c\right)}\)

2,cho x,y,z thỏa mãn x+y+z=5 và xy+yz+xz=8 chứng minh rằng \(1\le x\le\frac{7}{3}\)

3, cho a,b,c>0 chứng minh rằng\(\frac{a^2}{2a^2+\left(b+c-a\right)^2}+\frac{b^2}{2b^2+\left(b+c-a\right)^2}+\frac{c^2}{2c^2+\left(b+a-c\right)^2}\le1\)

4,cho a,b,c là các số thực bất kỳ chứng minh rằng \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\left(ab+bc+ac-1\right)^2\)

5, cho a,b,c > 1 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)chứng minh rằng \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{a+b+c}\)

Cho \(a;b;c\ge0\) và \(a+b+c=1\)

Tìm GTLN của \(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

                                                 Bài làm:

Ta có: \(a;b;c\ge0\) và \(a+b+c=1\)

Áp dụng Bđt Cô-si cho 2 số không âm, ta có: 

 \(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

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

\(\le\sqrt{\frac{3}{2}}\left(\frac{a+b+\frac{2}{3}}{2}+\frac{b+c+\frac{2}{3}}{2}+\frac{c+a+\frac{2}{3}}{2}\right)\)\(=\sqrt{\frac{3}{2}}.2=\sqrt{6}\)

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