ta có P=\(\frac{x^2}{x\sqrt{y+3}}+\frac{y^2}{y\sqrt{z+3}}+\frac{z^2}{z\sqrt{x+3}}\ge\frac{\left(x+y+z\right)^2}{x\sqrt{y+3}+y\sqrt{z+3}+z\sqrt{x+3}}\)

mà \(\left(x\sqrt{y+3}+...\right)^2\le\left(x+y+z\right)\left(xy+yz+zx+3x+3y+3z\right)\le3\left(9+3\right)=36\) ( vì xy+yz+zx<=3)

=>\(x\sqrt{y+3}+...\le6\Rightarrow P\ge\frac{9}{6}=\frac{3}{2}\)

dấu = xảy ra <=> x=y=z=1

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

đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)

\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)

\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)

\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)

\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)

\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)

\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)

\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)

Đề bài: ax,y,z >0 và \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\). Tìm Min P= \(\dfrac{x^3}{y+z}+\dfrac{y^3}{z+x}+\dfrac{z^3}{x+y}\).

ĐÁP ÁN:

Ta có: \(\dfrac{x^3}{y+z}+\dfrac{y+z}{36}+\dfrac{1}{162}+\dfrac{y^3}{x+z}+\dfrac{x+z}{36}+\dfrac{1}{162}+\dfrac{z^3}{x+y}+\dfrac{x+y}{36}+\dfrac{1}{162}\ge3\sqrt[3]{\dfrac{x^3}{y+z}.\dfrac{y+z}{36}.\dfrac{1}{162}}+3\sqrt[3]{\dfrac{y^3}{x+z}.\dfrac{x+z}{36}.\dfrac{1}{162}}+3\sqrt[3]{\dfrac{z^3}{x+y}.\dfrac{x+y}{36}.\dfrac{1}{162}}=3\sqrt[3]{\dfrac{x^3}{36.162}}+3\sqrt[3]{\dfrac{y^3}{36.162}}+3\sqrt[3]{\dfrac{z^3}{36.162}}=\dfrac{x+y+z}{6}.\)

=> P+\(\dfrac{x+y+z}{18}+\dfrac{1}{54}\)≥\(\dfrac{x+y+z}{6}\) <=> P≥\(\dfrac{x+y+z}{6}-\dfrac{x+y+z}{18}-\dfrac{1}{54}\)=\(\dfrac{x+y+z}{9}-\dfrac{1}{54}\)

Ta c/m đc: 3(x+y+z)≥(\(\sqrt{x}+\sqrt{y}+\sqrt{z}\))² <=> 2(x+y+z) ≥2\(\left(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\right)\)<=> x+y+z≥\(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\)(luôn đúng)

➩x+y+z ≥ \(\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^3}{3}=\dfrac{1}{3}\) => P≥\(\dfrac{1}{54}\). Dấu ''='' xảy ra <=> x=y=z=\(\dfrac{1}{9}\)