Cho a, b, c, d > 0 và a+b+c+d=4. Chứng minh rằng:
[tex]\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq 2[/tex]
Giúp mình nha.
Ta có: [tex]\frac{a}{1+b^2c}=a-\frac{ab^2c}{1+b^2c}\geq a-\frac{ab^2c}{2b\sqrt{c}}=a-\frac{ab\sqrt{c}}{2}[/tex]
Lại có: [tex]a-\frac{ab\sqrt{c}}{2}=a-\frac{b\sqrt{a.ac}}{2}\geq a-\frac{1}{4}b(a+ac)=a-\frac{1}{4}(ab+abc)[/tex]
Vậy [tex]\frac{a}{1+b^2c}\geq a-\frac{1}{4}(ab+abc)[/tex]
Chứng minh tương tự:
[tex]\frac{b}{1+c^2d}\geq b-\frac{1}{4}(bc+bcd) \\ \frac{c}{1+d^2a}\geq c-\frac{1}{4}(cd+cda) \\ \frac{d}{1+a^2b}\geq d-\frac{1}{4}(da+dab)[/tex]
[tex]\Rightarrow \frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq a+b+c+d-\frac{1}{4}\left [ (ab+bc+cd+da)+abc+bcd+cda+dab \right ][/tex]
Lại có: [tex]ab+bc+cd+da=(a+c)(b+d)\leq \left ( \frac{a+b+c+d}{2} \right )^2=4 \\ abc+bcd+cda+dab=abcd\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \right )\leq \left ( \frac{a+b+c+d}{4} \right )^4.\frac{16}{a+b+c+d}=\frac{1}{16}(a+b+c+d)^3=4[/tex]
Do đó, [tex]\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq a+b+c+d-2=2[/tex]
Dấu = xảy ra [tex]\Leftrightarrow a=b=c=d=1[/tex]
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