Đề có thiếu gì không bạn?

Áp dụng BĐT Cô-si cho 4 số \(a^3;1\) số \(b^3\) và \(1\) số \(c^3\) ta có:

\(4a^3+b^3+c^3\ge6\sqrt[6]{a^{12}.b^3.c^3}=6a^2\sqrt{bc}\left(1\right)\)

Tương tự như trên ta có:

\(4b^3+c^3+a^3\ge6b^2\sqrt{ca}\left(2\right)\)

\(4c^3+a^3+b^3\ge6c^2\sqrt{ab}\left(3\right)\)

Cộng theo vế các BĐT \(\left(1\right)\left(2\right)\)\(\left(3\right)\) ta được:

\(6\left(a^3+b^3+c^3\right)\ge6\left(a^2\sqrt{bc}+b^2\sqrt{ca}+c^2\sqrt{ab}\right)\)

\(\Leftrightarrow a^3+b^3+c^3\ge a^2\sqrt{bc}+b^2\sqrt{ca}+c^2\sqrt{ab}\left(đpcm\right)\)

\(\left(???\right)\)

\(VT=\sqrt{a^3}.\sqrt{abc}+\sqrt{b^3}.\sqrt{abc}+\sqrt{c^3}.\sqrt{abc}\)

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

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

\(VT\le a^3+b^3+c^3\)

Dấu "=" xảy ra khi \(a=b=c\)

Ta có: \(3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2=\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\)

\(\Rightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\) nên với \(x,y,z>0\) ta có:

\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\) áp dụng ta có:

\(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\le\sqrt{3\left(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\right)}\)

Với: \(x,y>0\) ta có: \(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Áp dụng ta được:

 \(\frac{1}{ab+a+2}=\frac{1}{ab+1+a+1}=\frac{1}{ab+abc+a+1}=\frac{1}{ab\left(c+1\right)+\left(a+1\right)}\)

\(\le\frac{1}{4}\left(\frac{1}{ab\left(c+1\right)}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{abc}{ab\left(c+1\right)}+\frac{1}{a+1}\right)=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)

Vậy ta có: \(\frac{1}{ab+a+2}\le\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)

Tương tự như trên ta có: \(\frac{1}{bc+b+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{b+1}\right)\) và \(\frac{1}{ca+c+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\) nên:

\(\Rightarrow\sqrt{3\left(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\right)}\)

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

Vậy \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\le\frac{3}{2}\left(đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

Đặt \(\left(a;b;c\right)=\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\). BĐT quy về:\(\Sigma_{cyc}\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}\le\frac{3}{2}\)

Áp dụng liên hoàn BĐT Cô si:

\(VT=\Sigma_{cyc}\sqrt{\frac{yz}{\left(xy+yz\right)+\left(xz+yz\right)}}\le\Sigma_{cyc}\sqrt{\frac{yz}{4}\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)}\)

\(=\frac{1}{2}\Sigma_{cyc}\sqrt{1\left(\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)}\le\frac{1}{4}\Sigma_{cyc}\left(1+\frac{yz}{xy+yz}+\frac{yz}{xz+yz}\right)=\frac{3}{2}\)

Áp dụng bất đẳng thức Cô - si ta có:

 \(\frac{1}{\sqrt{ab+a+2}}\le\left(\frac{1}{4}+\frac{1}{ab+a+2}\right)\)

Tương tự:

=> \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}\)

\(\le\frac{3}{4}+\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\)(1)

Áp dụng:  \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) với x, y >0 

Ta có: \(\frac{1}{ab+a+2}=\frac{1}{\frac{ab}{abc}+a+2}\le\frac{1}{4}.\left(\frac{1}{\frac{1}{c}+1}+\frac{1}{a+1}\right)\)vì abc =1

\(=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)\)

Tương tự 

=> \(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\)

\(\le\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{a+1}\right)+\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{b+1}\right)+\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\)

\(=\frac{1}{4}\left(\frac{c}{c+1}+\frac{1}{c+1}+\frac{1}{a+1}+\frac{a}{a+1}+\frac{1}{b+1}+\frac{b}{b+1}\right)\)

\(=\frac{1}{4}\left(1+1+1\right)=\frac{3}{4}\)(2)

Từ (1); (2) 

=> \(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ac+c+2}}\le\frac{3}{4}+\frac{3}{4}=\frac{3}{2}\)

Dấu "=" xảy ra <=> a = b = c = 1

\(ab+bc+ca\le1\)

\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)

\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)

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

ta có:

\(c+ab=c.1+ab=c\left(a+b+c\right)+ab=ca+cb+c^2+ab=\left(c+a\right)\left(c+b\right)\)

tương tự như vậy thì \(P=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(c+a\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)

áp dụng bđt cô si ta có:

\(\frac{a}{a+c}+\frac{b}{b+c}\ge2\sqrt{\frac{ab}{\left(c+a\right)\left(b+c\right)}};\frac{b}{a+b}+\frac{c}{c+a}\ge2\sqrt{\frac{bc}{\left(a+b\right)\left(c+a\right)}};\frac{a}{a+b}+\frac{c}{b+c}\ge2\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)

\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+a}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}\right)=\frac{3}{2}\left(Q.E.D\right)\)

Ta có:\(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a\left(a+b\right)+c\left(a+b\right)}}\)

\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) (Áp dụng BĐT AM-GM)

Tương tự với hai BĐT còn lại và cộng theo vế ta thu được đpcm.