giúp mk với :(((((((((((
\(5\left(x^2+y^2+z^2\right)=9\left(xy+2yz+zx\right)\\ \Leftrightarrow5x^2+5\left(y^2+z^2\right)-9x\left(y+z\right)-18yz=0\\ \Leftrightarrow5x^2-9x\left(y+z\right)=18yz-5\left(y^2+z^2\right)\)
Ta có \(yz\le\dfrac{\left(y+z\right)^2}{4};y^2+z^2\ge\dfrac{\left(y+z\right)^2}{2}\)
\(\Leftrightarrow18yz-5\left(y^2+z^2\right)\le\dfrac{9\left(y+z\right)^2}{2}-\dfrac{5\left(y+z\right)^2}{2}=2\left(y+z\right)^2\\ \Leftrightarrow5x^2-9x\left(y+z\right)\le2\left(y+z\right)^2\\ \Leftrightarrow5x^2-9x\left(y+z\right)-2\left(y+z\right)^2\le0\\ \Leftrightarrow\left[x-2\left(y+z\right)\right]\left(5x+y+z\right)\le0\\ \Leftrightarrow x-2\left(y+z\right)\le0\left(x,y,z>0\right)\\ \Leftrightarrow x\le2\left(y+z\right)\)
Ta có \(P=\dfrac{x}{y^2+z^2}-\dfrac{1}{\left(x+y+z\right)^3}\le\dfrac{x}{\dfrac{\left(y+z\right)^2}{2}}-\dfrac{1}{\left(x+y+z\right)^3}\)
\(\Leftrightarrow P\le\dfrac{2x}{\left(y+z\right)^2}-\dfrac{1}{\left[2\left(y+z\right)+y+z\right]^3}\le\dfrac{4\left(y+z\right)}{\left(y+z\right)^2}-\dfrac{1}{27\left(y+z\right)^3}\\ \Leftrightarrow P\le\dfrac{4}{y+z}-\dfrac{1}{27\left(y+z\right)^3}\)
Đặt \(t=\dfrac{1}{y+z}>0\Leftrightarrow P\le4t-\dfrac{t^3}{27}\)
\(\Leftrightarrow16-P\ge16-4t+\dfrac{t^3}{27}=\dfrac{t^3-108t+432}{27}\\ \Leftrightarrow16-P\ge\dfrac{\left(t+12\right)\left(t-6\right)^2}{27}\ge0\left(t>0\right)\\ \Leftrightarrow P\le16\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{3};y=z=\dfrac{1}{12}\)
