Question

設(shè)函數(shù)（x＞1）．
（I）求函數(shù)f（x）的最小值；
（Ⅱ）若m，t∈R⁺，且，求證：；
（Ⅲ）若，且，求證：．

Answer 1

（I）解：求導數(shù)可得：f′(x)=1/x²log_{2}(x-1)（x＞1）令f′（x）≥0，得x≥2，所以f（x）在（1，2）上遞減，在（2，+∞）上遞增．所以f（x）_min=f（2）=-1．（Ⅱ）證明：log_{2}m/m+log_{2}t/t=log_{2}m/m-log_{2}{1/t}{t}=log_{2}m/m-（1-1/m）log_{2}(1-1/m)=-[m-1/mlog_{2}(m-1)-log_{2}m]由（I）知當x＞1時，x-1/xlog_{2}(x-1)-log_{2}x≥-1，又m，t∈R⁺，且1/m+1/t=1，∴m＞1∴m-1/mlog_{2}(m-1)-log_{2}m≥-1∴l(xiāng)og_{2}m/m+log_{2}t/t≤1∴tlo{g}{^{ }_{2}}m+mlo{g}{^{ }_{2}}t≤mt．（Ⅲ）證明：用數(shù)學歸納法證明如下：1°當n=1時，由（Ⅱ）可知，不等式成立；2°假設(shè)n=k時不等式成立，即若a_{1}，a_{2}，a_{3}，…，a_{2^{k}}∈R^{+}，且1/a_{1}+1/a_{2}+1/a_{3}+…+1/a_{2^{k}}=1時，不等式lo{g/^{ _{2}}a_{1}}{a_{1}}+lo{g/^{ _{2}}a_{2}}{a_{2}}+lo{g/^{ _{2}}a_{3}}{a_{3}}+…+lo{g/^{ _{2}}a_{2^{k}}}{a_{2^{k}}}≤k成立現(xiàn)需證當n=k+1時不等式也成立，即證：若a_{1}，a_{2}，a_{3}，…，a_{2^{k+1}}∈R^{+}，且1/a_{1}+1/a_{2}+1/a_{3}+…+1/a_{2^{k+1}}=1時，不等式lo{g/^{ _{2}}a_{1}}{a_{1}}+lo{g/^{ _{2}}a_{2}}{a_{2}}+lo{g/^{ _{2}}a_{3}}{a_{3}}+…+lo{g/^{ _{2}}a_{2^{k+1}}}{a_{2^{k+1}}}≤k+1成立．證明如下：設(shè)1/a_{1}+1/a_{2}+1/a_{3}+…+1/a_{2^{k}}=x，則1/xa_{1}+1/xa_{2}+1/xa_{3}+…+1/xa_{2^{k}}=1∴l(xiāng)o{g/^{ _{2}}xa_{1}}{xa_{1}}+lo{g/^{ _{2}}xa_{2}}{xa_{2}}+lo{g/^{ _{2}}xa_{3}}{xa_{3}}+…+lo{g/^{ _{2}}xa_{2^{k}}}{xa_{2^{k}}}≤k∴-lo{g/^{ _{2}}xa_{1}}{a_{1}}+-lo{g/^{ _{2}}xa_{2}}{a_{2}}+-lo{g/^{ _{2}}xa_{3}}{a_{3}}+…+-lo{g/^{ _{2}}xa_{2^{k}}}{a_{2^{k}}}≥-kx∴-lo{g/^{ _{2}}a_{1}}{a_{1}}+-lo{g/^{ _{2}}a_{2}}{a_{2}}+-lo{g/^{ _{2}}a_{3}}{a_{3}}+…+-lo{g/^{ _{2}}a_{2^{k}}}{a_{2^{k}}}≥-kx+xlog₂x…①同理-lo{g/^{ _{2}}a_{2^{k}+1}}{a_{2^{k}+1}}+-lo{g/^{ _{2}}a_{2^{k}+2}}{a_{2^{k}+2}}+…+-lo{g/^{ _{2}}a_{2^{k+1}}}{a_{2^{k+1}}}≥-k(1-x)+（1-x）log₂（1-x）…②由①+②得：-lo{g/^{ _{2}}a_{1}}{a_{1}}+-lo{g/^{ _{2}}a_{2}}{a_{2}}+-lo{g/^{ _{2}}a_{3}}{a_{3}}+…+-lo{g/^{ _{2}}a_{2^{k+1}}}{a_{2^{k+1}}}≥-k+[xlog₂x+（1-x）log₂（1-x）]又由（Ⅱ）令1/m=x，則1/t=1-x，其中∈x（0，1），則有l(wèi)og_{2}m/m+log_{2}t/t≤1∴xlog₂x+（1-x）log₂（1-x）≥-1∴-k+[xlog₂x+（1-x）log₂（1-x）]≥-k-1∴l(xiāng)o{g/^{ _{2}}a_{1}}{a_{1}}+lo{g/^{ _{2}}a_{2}}{a_{2}}+lo{g/^{ _{2}}a_{3}}{a_{3}}+…+lo{g/^{ _{2}}a_{2^{k+1}}}{a_{2^{k+1}}}≤k+1∴當n=k+1時，原不等式也成立．綜上，由1°和2°可知，原不等式均成立．