Lời giải.
Theo bất đẳng thức `Cô-si` ta có:
`ab\sqrt{c}=a\sqrt{b.bc}≤{a(b+bc)}/2`
Tương tự: `bc\sqrt{d}≤{b(c+cd)}/2,cd\sqrt{a}≤{c(d+da)}/2,ad\sqrt{b}≤{d(a+ad)}/2`
`=>ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+ad\sqrt{b}`
`≤{ab+bc+cd+da+abc+bcd+cda+dab}/2`
Ta có:
`ab+bc+cd+da=(ad+da)+(bc+cd)=(a+c)(b+d)≤{(a+b+c+d)^2}/4=4^2/4=4`
`=>{ab+bc+cd+da}/2≤4/2=2`
Lại có:
`abc+bcd+cda+dab=(abc+bcd)+(cda+dab)=bc(a+d)+da(c+b)`
Mà `bc≤{(b+c)^2}/4,da≤{(d+a)^2}/4`
`=>bc(a+d)+da(c+d)≤{(a+d)(b+c)^2}/4+{(c+b)(d+a)^2}/4` `(1)`
Đặt `a+d=x(x>0),c+b=y(y>0)=>x+y=4` thì `(1)<=> {xy^2+yx^2}/4`
`={xy(x+y)}/4`
Lại có: `xy≤{(x+y)^2}/4`
`=>{xy(x+y)}/4≤{(x+y)^3}/16=4^3/16=4`
`<=>abc+bcd+cda+dab≤4`
`<=>{abc+bcd+cda+dab}/2≤2`
Cộng vế, ta suy ra `{ab+bc+cd+da+abc+bcd+cda+dab}/2≤2+2=4.`
Mà `ab\sqrt{c}+bc\sqrt{d}+cd\sqrt{a}+ad\sqrt{b}`
`≤{ab+bc+cd+da+abc+bcd+cda+dab}/2`
`=>dpcm.`
Dấu "=" xảy ra `<=>a=b=c=d=1.`