Ta chứng minh BĐT sau cho các số dương:

\(x^5+y^5\ge xy\left(x^3+y^3\right)\)

\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)

Áp dụng:

\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)

Tương tự và cộng lại:

\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)

\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)

\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz và AM-GM:

$M=\frac{b^2+c^2}{a^2}+a^2(\frac{1}{b^2}+\frac{1}{c^2})$

$\geq \frac{b^2+c^2}{a^2}+a^2.\frac{4}{b^2+c^2}$

$=(\frac{b^2+c^2}{a^2}+\frac{a^2}{b^2+c^2})+\frac{3a^2}{b^2+c^2}$

$\geq \sqrt{\frac{b^2+c^2}{a^2}.\frac{a^2}{b^2+c^2}}+\frac{3(b^2+c^2)}{b^2+c^2}$

$=2+3=5$

Vậy $M_{\min}=5$ 

Theo bđt Cauchy - Schwart ta có:

\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)

\(=\frac{\left(ab+bc+ca\right)^2}{abc\left(ab+bc+ca\right)+3a^2b^2c^2}\)

Đặt \(ab+bc+ca=x;abc=y\).

Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)

\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )

Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1

sai rồi nhé bạn 

làm sao mà \(x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\)lại luôn đúng

Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).

1)

   +)  Ta có

            \(\left(a-b\right)^2\ge0\)

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

        \(\Leftrightarrow a^2+b^2\ge2ab\)

        \(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab\)

        \(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

        \(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)  ( đpcm )

     + )   Theo phần trên

             \(a^2+b^2\ge2ab\)

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

           \(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

            \(\Leftrightarrow ab\le\frac{1}{4}\left(a+b\right)^2\)  ( đpcm )

2, 

Ta có: \(5\left(x^2+y^2+z^2\right)-9x\left(y+z\right)-18yz=0\Leftrightarrow5x^2-9x\left(y+z\right)+5\left(y+z\right)^2=28yz\le7\left(y+z\right)^2\)\(\Leftrightarrow5x^2-9x\left(y+z\right)-2\left(y+z\right)^2\le0\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-9.\frac{x}{y+z}-2\le0\)\(\Leftrightarrow\left(5.\frac{x}{y+z}+1\right)\left(\frac{x}{y+z}-2\right)\le0\Leftrightarrow\frac{x}{y+z}\le2\)(Do \(5.\frac{x}{y+z}+1>0\forall x,y,z>0\))

\(\Rightarrow E=\frac{2x-y-z}{y+z}=2.\frac{x}{y+z}-1\le2.2-1=3\)

Đẳng thức xảy ra khi \(y=z=\frac{x}{4}\)

:v

Với mọi x;y dương, ta có:

\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow2x^2+2y^2\ge x^2+y^2+2xy\)

\(\Leftrightarrow x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\)

Đồng thời \(x^2+y^2\ge2xy\Rightarrow x^2+y^2+2xy\ge4xy\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Áp dụng: đặt vế trái của BĐT cần chứng minh là P, ta có:

\(P=\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{1}{2}\left(a+\dfrac{1}{b}+b+\dfrac{1}{a}\right)^2=\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)

\(P\ge\dfrac{1}{2}\left(a+b+\dfrac{4}{a+b}\right)^2=\dfrac{1}{2}\left(3+\dfrac{4}{3}\right)^2=\dfrac{169}{18}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{3}{2}\)

\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)

\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)